[OI]省选前模板整理

 

省选前把板子整理一遍,如果发现有脑抽写错的情况,欢迎各位神犇打脸 :)

 

数学知识

 

数论:

//组合数
//C(n,m) 在n个数中选m个的方案数 
ll C[N][N];
void get_C(int n)
{
	for(int i=1;i<=n;i++) 
	{
		C[i][i]=C[i][0]=1;
		for(int j=1;j<i;j++)
			C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
	}
}
//欧几里得算法
//(a,b) 
ll gcd(ll a,ll b)
{
	return b==0? a:gcd(b,a%b);
}
//拓展欧几里得算法
//解同余方程 a*x+b*y = (a,b) 
ll exgcd(ll a,ll b,ll& d,ll& x,ll& y)
{
	if(!b) { d=a; x=1; y=0; }
	else { exgcd(b,a%b,d,y,x); y-=x*(a/b); }
}
//逆元 
//a*inv(a,n) = 1 mod n
ll inv(ll a,ll n)
{
	ll d,x,y;
	exgcd(a,n,d,x,y);
	return d==1? (x+n)%n:-1;
}
//lucas定理
//计算较大,有模数的组合数 
ll fac[N];
void get_pre(int n)
{
	for(int i=1;i<=n;i++) 
		fac[i]=(fac[i-1]*i)%mod;
}
ll C(ll n,ll m,ll mod)
{
	if(n<m) return 0;
	if(n<mod&&m<mod) 
		return fac[n]*inv(fac[m],mod)%mod*inv(fac[n-m],mod)%mod;
	return C(n/mod,m/mod,mod)*C(n%mod,m%mod,mod)%mod;
}
//快速幂
//a^p % mod 
ll pow(ll a,ll p,ll mod)
{
	ll ans=1;
	while(p) 
	{
		if(p&1) ans=(ans*a)%mod;
		a=(a*a)%mod; p>>=1;
	}
	return ans;
}
//中国剩余定理
//解线性同余方程组 
//sigma{ ai*(1-ai*mi) } % M  , ai*mi+wi*y=1
ll a[N],m[N];
ll china(int n)
{
	ll M=1,d,x=0,y;
	for(int i=1;i<=n;i++) M*=m[i];
	for(int i=1;i<=n;i++)
	{
		ll w=M/m[i];
		exgcd(m[i],w,d,d,y);
		x=(x+y*w*a[i])%M;
	}
	return (x+M)%M;
}
//大步小步算法
//计算a^x=b mod n中的最小x
map<int,int> mp;
int BSGS(int a,int b,int n)
{
	int m=sqrt(n)+1,e=1,i;
	int v=inv(pow(a,m,n),n);
	mp[e]=0;
	for(i=1;i<m;i++)
	{
		e=(e*m)%n;
		if(!mp.count(e)) mp[e]=i;
	}
	for(i=0;i<m;i++)
	{
		if(mp.count(b)) return i*m+mp[b];
		b=(b*v)%mod;
	}
	return -1;
}
//快速筛法求素数表
int su[N],vis[N];
void get_su(int n)
{
	for(int i=2;i<=n;i++)
	{
		if(!vis[i]) su[++su[0]]=i;
		for(int j=1;j<=su[0]&&i*su[j]<=n;j++)
		{
			vis[i*su[j]]=1;
			if(i%su[j]==0) break;
		}
	}
}
//欧拉函数
//phi(n)小于n的数中与n互素的数的个数 
ll get_phi(int n)
{
	int m=sqrt(n)+1;
	ll ans=n;
	for(int i=2;i<=m;i++) if(n%i==0) 
	{
		ans=ans/i*(i-1);
		while(n%i==0) n/=i;
	}
	if(n>1) ans=ans/n*(n-1);
	return ans;
}
ll phi[N];
void get_phi_table(int n)
{
	phi[1]=1;
	for(int i=2;i<=n;i++) if(!phi[i])
	{
		for(int j=i;j<=n;j+=i) 
		{
			if(!phi[j]) phi[j]=j;
			phi[j]=phi[j]/i*(i-1);
		}
	}
}
//莫比乌斯函数 int mu[N],su[N],vis[N]; void get_mu(int n) { mu[1]=1; for(int i=2;i<=n;i++) { if(!vis[i]) mu[i]=-1,su[++su[0]]=i; for(int j=1;j<=su[0]&&i*su[j]<=n;j++) { vis[i*su[j]]=1; if(i%su[j]==0) mu[i*su[j]]=0; else mu[i*su[j]]=-mu[i]; } } } //高斯消元 //解线性方程组 double a[N][N]; void gause(int n) { for(int i=1;i<=n;i++) { int r=i; for(int j=i+1;j<=n;j++) if(fabs(a[j][i])>fabs(a[r][i])) r=i; for(int j=1;j<=n+1;j++) swap(a[i][j],a[r][j]); for(int j=n+1;j>=i;j--) for(int k=i+1;k<=n;k++) a[k][j]-=a[k][i]/a[i][i]*a[i][j]; } for(int i=n;i;i--) { for(int j=i+1;j<=n;j++) a[i][n+1]-=a[j][n+1]*a[i][j]; a[i][n+1]/=a[i][i]; } }

 

高精度:

int trans(char* s,int st,int ed)
{
    int x=0;
    for(int i=st;i<ed;i++) x=x*10+s[i]-'0';
    return x;
}

struct Bign
{
    int len; ll N[maxn];
    Bign() { len=0; memset(N,0,sizeof(N)); }
    Bign(ll num) {  *this=num; }
    Bign(const char* s) { *this=s; }
    void print()
    {
        printf("%d",N[len-1]);
        for(int i=len-2;i>=0;i--)
            printf("%08d",N[i]);
        puts("");
    }
    Bign operator = (const ll x)
    {
        ll num=x;
        while(num>base)
        {
            N[len++]=num%base;
            num/=base;
        }
        if(num) N[len++]=num;
        return *this;
    }
    Bign operator = (char* s)
    {
        int L=strlen(s);
        len=(L-1)/wlen+1;
        for(int i=0;i<len;i++)
        {
            int ed=L-i*wlen;
            int st=max(0,ed-wlen);
            N[i]=trans(s,st,ed);
        }
        return *this;
    }
    bool operator < (const Bign& B) const
    {
        if(len!=B.len) return len<B.len;
        for(int i=len-1;i>=0;i--)
            if(N[i]!=B.N[i]) return N[i]<B.N[i];
        return 0;
    }
    bool operator <= (const Bign& B) const
    {
        return !(B<(*this));
    }
     
    void clear()
    {
        while(len>1&&N[len-1]==0) len--;
    }
     
    Bign operator + (const Bign& B) const
    {
        Bign C;
        C.len=max(len,B.len)+10;
        for(int i=0;i<C.len;i++)
        {
            C.N[i]+=N[i]+B.N[i];
            C.N[i+1]+=C.N[i]/base;
            C.N[i]%=base;
        }
        C.clear();
        return C;
    }
    Bign operator - (Bign B)
    {
        Bign C=*this;
        C.len=max(C.len,B.len);
        for(int i=0;i<C.len;i++)
        {
            if(C.N[i]<B.N[i]) C.N[i+1]--,C.N[i]+=base;
            C.N[i]=C.N[i]-B.N[i];
        }
        C.clear();
        return C;
    }
    Bign operator * (const Bign& B) const
    {
        Bign C;
        C.len=len+B.len;
        for(int i=0;i<len;i++)
            for(int j=0;j<B.len;j++)
                C.N[i+j]+=N[i]*B.N[j];
        for(int i=0;i<C.len;i++)
        {
            C.N[i+1]+=C.N[i]/base;
            C.N[i]%=base;
        }
        C.clear();
        return C;
    }
    Bign operator / (const Bign& B)
    {
        Bign C,F;
        C.len=len;
        for(int i=len-1;i>=0;i--)
        {
            F=F*base;
            F.N[0]=N[i];
            while(B<=F)
            {
                F=F-B;
                C.N[i]++;
            }
        }
        C.clear();
        return C;
    }
    Bign operator % (const Bign& B)
    {
        Bign r=*this/B;
        return *this-r*B;
    }
 
}A,B;

  

 

矩阵乘法:

//矩阵乘法 
struct Mat 
{
	int r,c; ll N[maxn][maxn];
	Mat(int r=0,int c=0) 
	{
		this->r=r,this->c=c;
		memset(N,0,sizeof(N));
	}
	Mat operator * (const Mat& B) const
	{
		Mat C(r,B.c);
		for(int i=0;i<r;i++)
			for(int j=0;j<B.c;j++)
				for(int k=0;k<c;k++)
					C.N[i][j]=(C.N[i][j]+N[i][k]*B.N[k][j])%mod;
		return C;
	}
	Mat operator ^ (int p)
	{
		Mat ans(r,r),tmp=*this;
		for(int i=0;i<r;i++) ans.N[i][i]=1;
		while(p)
		{
			if(p&1) ans=ans*tmp;
			tmp=tmp*tmp; p>>=1;
		}
		return ans;
	}
	
}; 

 

 

数据结构

 

树状数组:

//树状数组
int C[N],mx;
void Add(int x,int v)
{
    for(int i=x;i<=mx;i+=i&-i) C[i]+=v;
}
int query(int x)
{
    int ans=0;
    for(int i=x;i;i-=i&-i) ans+=C[i];
    return ans;
}

  

线段树:

//线段树
//区间加,区间乘,区间求和 
int mod;
struct Tnode 
{
	int u,l,r;
	ll sum,add,mul;
	void mulv(ll x) {
		sum=(sum*x)%mod;
		mul=(mul*x)%mod;
		add=(add*x)%mod;
	}
	void addv(ll x) {
		sum=(sum+(r-l+1)*x%mod)%mod;
		add=(add+x)%mod;
	}
	void pushdown() ;
	void maintain() ;
}T[N];
void Tnode::pushdown() {
	if(mul^1) {
		T[u<<1].mulv(mul);
		T[u<<1|1].mulv(mul);
		mul=1;
	}
	if(add) {
		T[u<<1].addv(add);
		T[u<<1|1].addv(add);
		add=0;
	}
}
void Tnode::maintain() {
	sum=(T[u<<1].sum+T[u<<1|1].sum)%mod;
}

void update(int u,int L,int R,int x,int f)
{
	T[u].pushdown();
	if(L<=T[u].l&&T[u].r<=R) {
		if(!f) T[u].addv(x);
		else T[u].mulv(x);
	} else {
		int mid=T[u].l+T[u].r>>1;
		if(L<=mid) update(u<<1,L,R,x,f);
		if(mid<R) update(u<<1|1,L,R,x,f);
		T[u].maintain();
	}
}
ll query(int u,int L,int R)
{
	T[u].pushdown();
	if(L<=T[u].l&&T[u].r<=R)
		return T[u].sum;
	else {
		int mid=T[u].l+T[u].r>>1;
		ll ans=0;
		if(L<=mid) ans=(ans+query(u<<1,L,R))%mod;
		if(mid<R) ans=(ans+query(u<<1|1,L,R))%mod;
		return ans; 
	}
}
ll a[N];

void build(int u,int l,int r)
{
	T[u]=(Tnode){ u,l,r,0,0,1 };
	if(l==r) {
		T[u].sum=a[l];
	} else {
		int mid=l+r>>1;
		build(u<<1,l,mid);
		build(u<<1|1,mid+1,r);
		T[u].maintain();
	}
}

 

Treap:

//Treap 
struct Node 
{
	Node *ch[2];
	int v,r,m,w,s;
	Node(int v):v(v) { ch[0]=ch[1]=NULL; r=rand(); s=w=1; }
	int cmp(int x) {
		if(v==x) return -1;
		return x<v? 0:1;
	}
	void maintain() {
		s=w;
		if(ch[0]!=NULL) s+=ch[0]->s;
		if(ch[1]!=NULL) s+=ch[1]->s;
	}
};

void rotate(Node* &o,int d)
{
	Node* k=o->ch[d^1];o->ch[d^1]=k->ch[d];k->ch[d]=o;
	o->maintain(); k->maintain(); o=k;
}
void insert(Node *&o,int x)
{
	if(o==NULL) o=new Node(x);
	int d=o->cmp(x);
	if(d==-1) o->w++;
	else {
		insert(o->ch[d],x);
		if(o->ch[d]->r > o->r) rotate(o,d^1);
	}
	o->maintain();
}
void remove(Node *&o,int x)
{
	int d=o->cmp(x);
	if(d==-1) 
	{
		if(o->s>1) { o->w--; o->maintain(); return ; }
		else {
			if(o->ch[0]!=NULL&&o->ch[1]!=NULL) {
				int d2=o->ch[0]->r > o->ch[1]->r ? 1:0;
				rotate(o,d2); remove(o->ch[d2],x);
			} else {
				if(o->ch[0]!=NULL) o=o->ch[0]; else o=o->ch[1];
				delete o;
			}
		}
	} else 
		remove(o->ch[d],x);
	if(o!=NULL) o->maintain();
}
int kth(Node* o,int rk)
{
	if(o==NULL) return 0;
	int s=o->ch[0]==NULL? 0:o->ch[0]->s;
	if(rk==s+1) return o->v;
	else if(rk<=s) return kth(o->ch[0],rk);
	else return kth(o->ch[1],rk-s-o->w);
}
int rank(Node* o,int x)
{
	if(o==NULL) return 0;
	int s=o->ch[0]==NULL? 0:o->ch[0]->s;
	int d=o->cmp(x);
	if(d==-1) return 1;
	else if(d==0) return rank(o->ch[0],x);
	else return s+o->w+rank(o->ch[1],x);
}
int tmp;
void before(Node* o,int x)
{
	if(o==NULL) return ;
	if(o->v<x) { tmp=max(tmp,o->v); before(o->ch[1],x); }
	else before(o->ch[0],x);
}
void after(Node* o,int x)
{
	if(o==NULL) return ;
	if(o->v>x) { tmp=min(tmp,o->v); after(o->ch[0],x); }
	else after(o->ch[1],x);
}

 

splay:

//splay自上而下 
struct Node 
{
	Node *ch[2];
	int s;
	int cmp(int x)
	{
		int d=x-ch[0]->s;
		if(d==1) return -1;
		return d<=0? 0:1;
	}
	void maintain()
	{
		s=ch[0]->s+ch[1]->s;
	}
	void pushdown() {}
}mempool[N],*G=mempool;

Node* null=new Node();
void rotate(Node* &o,int d)
{
	Node* k=o->ch[d^1]; o->ch[d^1]=k->ch[d],k->ch[d]=o;
	o->maintain(); k->maintain(); o=k;
}
void splay(Node* &o,int k)
{
	o->pushdown();
	int d=o->cmp(k);
	if(d==1) k-=o->ch[0]->s+1;
	if(d!=-1) {
		Node* p=o->ch[d];
		p->pushdown();
		int d2=p->cmp(k),k2=d2==0? k:k-p->ch[d]->s-1;
		if(d2!=-1) {
			splay(p->ch[d2],k2);
			if(d==d2) rotate(o,d^1); else rotate(o->ch[d],d);
		}
		rotate(o,d^1);
	}
}
Node* merge(Node* left,Node* right)
{
	splay(left,left->s);
	left->ch[1]=right,left->maintain();
	return left;
}
void split(Node* o,int k,Node*&left,Node*&right)
{
	splay(o,k);
	left=o,right=left->ch[1],left->ch[1]=NULL;
	left->maintain();
}
Node* build(int l,int r)
{
	if(r<l) return null;
	int mid=l+r>>1;
	G->s=1;
	G->ch[0]=build(l,mid-1);
	G->ch[1]=build(mid+1,r);
	G->maintain();
	return G++;
}

 

主席树:

//主席树 
struct Tnode 
{
	Tnode *ls,*rs;
	int sum;
} *T[N*50],mempool[N*50],*G=mempool;

Tnode* Nw(Tnode* l,Tnode* r,int x)
{
	G->ls=l,G->rs=r,G->sum=x;
	return G++;
}
Tnode* build(Tnode* p,int l,int r,int pos)
{
	if(l==r) 
		return Nw(T[0],T[0],p->sum+1);
	else {
		int mid=l+r>>1;
		if(pos<=mid) return Nw(build(p->ls,l,mid,pos),p->rs,p->sum+1);
		else return Nw(p->ls,build(p->rs,mid+1,r,pos),p->sum+1);
	}
}
int query(Tnode* x,int l,int r,int pos)
{
	if(l==r) return x->sum;
	else {
		int mid=l+r>>1;
		if(pos<=mid) return query(x->ls,l,mid,pos);
		else return query(x->rs,mid+1,r,pos);
	}
}

 

Link-Cut-Tree

//Link-Cut-Tree 
namespace LCT 
{

	struct Node {
		Node *ch[2],*fa;
		int rev;
		//others v
		Node() {};
		Node(int x) ; 
		void reverse() {
			swap(ch[0],ch[1]);
			rev^=1;
		}
		void up_push() {
			if(fa->ch[0]==this||fa->ch[1]==this)
				fa->up_push();
			if(rev) {
				ch[0]->reverse();
				ch[1]->reverse();
				rev=0;
			}
		}
		void maintain() { }
	} T[N<<1],*null=&T[0];
	Node::Node(int x) {
		ch[0]=ch[1]=fa=null;
		rev=0; //v=x;
	}
	void rot(Node* o,int d) {
		Node* p=o->fa;
		p->ch[d]=o->ch[d^1];
		o->ch[d^1]->fa=p;
		o->ch[d^1]=p;
		o->fa=p->fa;
		if(p==p->fa->ch[0])
			p->fa->ch[0]=o;
		else if(p==p->fa->ch[1])
			p->fa->ch[1]=o;
		p->fa=o;
		p->maintain();
	}
	void splay(Node* o) {
		o->up_push();
		Node *nf,*nff;
		while(o->fa->ch[0]==o||o->fa->ch[1]==o) {
			nf=o->fa,nff=nf->fa;
			if(o==nf->ch[0]) {
				if(nf==nff->ch[0]) rot(nf,0);
				rot(o,0);
			} else {
				if(nf==nff->ch[1]) rot(nf,1);
				rot(o,1);
			}
		}
		o->maintain();
	}
	void Access(Node* o) {
		Node *son=null;
		while(o!=null) {
			splay(o);
			o->ch[1]=son;
			o->maintain();
			son=o; o=o->fa;
		}
	}
	void evert(Node* o) {
		Access(o); 
		splay(o);
		o->reverse();
	}
	void Link(Node* u,Node* v) {
		evert(u);
		u->fa=v;
	}
	void Cut(Node* u,Node* v) {
		evert(u);
		Access(v),splay(v);
		v->ch[0]=u->fa=null;
		v->maintain();
	}
	Node* find(Node* o) {
		while(o->fa!=null) o=o->fa;
		return o;
	}

}
using namespace LCT ;

 

 

 

2-SAT:

//2-sat
struct TwoSAT {
	int n;
	vector<int> g[N<<1];
	int st[N<<1],mark[N<<1],top;
	
	bool dfs(int x) {
		if(mark[x^1]) return 0;
		if(mark[x]) return 1;
		mark[x]=1;
		st[++top]=x;
		for(int i=0;i<g[x].size();i++)
			if(!dfs(g[x][i])) return 0;
		return 1;
	}
	void init(int n) {
		this->n=n;
		for(int i=0;i<2*n;i++) g[i].clear();
		memset(mark,0,sizeof(mark));
	}
	void addc(int x,int xval,int y,int yval) {
		x=x*2+xval;
		y=y*2+yval;
		g[x^1].push_back(y);
		g[y^1].push_back(x);
	}
	bool solve() {
		for(int i=0;i<2*n;i+=2) {
			if(!mark[i]&&!mark[i+1]) {
				top=0;
				if(!dfs(i)) {
					while(top) mark[st[top--]]=0;
					if(!dfs(i+1)) return 0;
				}
			}
		}
		return 1;
	}
	
} s;

  

有向图的强联通分量:

//tarjan求SCC 
struct Edge {
	int v,nxt;
}e[M];
int en=1,front[N];
void adde(int u,int v) 
{
	e[++en]=(Edge){v,front[u]}; front[u]=en;
}

int n,top,dfn;
int st[N],sccno[N],scc_cnt,pre[N],lowlink[N];

void tarjan(int u)
{
	pre[u]=lowlink[u]=++dfn;
	st[++top]=u;
	trav(u,i) {
		int v=e[i].v;
		if(!pre[v]) {
			tarjan(v);
			lowlink[u]=min(lowlink[u],lowlink[v]);
		} else 
		if(!sccno[v]) 
			lowlink[u]=min(lowlink[u],pre[v]);
	}
	if(lowlink[u]==pre[u]) {
		scc_cnt++;
		for(;;) {
			int x=st[top--];
			sccno[x]=scc_cnt;
			if(x==u) break;
		}
	}
}

 

无向图的边的双连通分量:

//BCC
struct Edge {
	int u,v,nxt;
}e[M];
int en=1,front[N];
void adde(int u,int v) 
{
	e[++en]=(Edge){u,v,front[u]}; front[u]=en;
}

Edge st[N];
vector<int> bcc[N];
int pre[N],iscut[N],bccno[N],top,dfn,bcc_cnt;

int dfs(int u,int fa)
{
	int lowu=pre[u]=++dfn;
	int child=0;
	trav(u,i) {
		int v=e[i].v;
		Edge E=e[i];
		if(!pre[v]) {
			st[++top]=E;
			child++;
			int lowv=dfs(v,u);
			lowu=min(lowu,lowv);
			if(lowv>=pre[u]) {
				iscut[u]=1;
				bcc_cnt++;
				for(;;) {
					Edge x=st[top--];
					if(bccno[x.u]!=bcc_cnt) {
						bccno[x.u]=bcc_cnt; 
						bcc[bcc_cnt].push_back(x.u);
					}
					if(bccno[x.v]!=bcc_cnt) {
						bccno[x.v]=bcc_cnt; 
						bcc[bcc_cnt].push_back(x.v);
					}
					if(x.u==u&&x.v==v) break;
				}
			}
		} else 
		if(pre[v]<pre[u] && v!=fa) {
			st[++top]=E;
			lowu=min(lowu,pre[v]);
		}
	}
	if(fa<0&&child==1) iscut[u]=0;
	return lowu;
}

  

最短路:

//spfa

struct Edge {
	int v,w,nxt;
}e[M];
int en=1,front[N];
void adde(int u,int v,int w) 
{
	e[++en]=(Edge){v,w,front[u]}; front[u]=en;
}

queue<int> q;
int inq[N],dis[N];
void spfa(int s)
{
	dis[s]=0; inq[s]=1;
	q.push(s);
	while(!q.empty()) {
		int u=q.front(); q.pop();
		inq[u]=0;
		trav(u,i) {
			int v=e[i].v;
			if(dis[v]>dis[u]+e[i].w) {
				dis[v]=dis[u]+e[i].w;
				if(!inq[v]) {
					inq[v]=1;
					q.push(v);
				}
			}
		}
	}
}

//dijkstra

struct Node {
    int id,dis;
    bool operator < (const Node& rhs) const
    {
        return dis>rhs.dis;
    }
};

priority_queue<Node> q;
int n,m,s;
int vis[N],dis[N];

void dijkstra(int s)
{
	FOR(i,1,n) dis[i]=inf;
    dis[s]=0; q.push((Node){s,0});
    while(!q.empty()) {
        int u=q.top().id;
        q.pop();
        if(vis[u]) continue;
        vis[u]=1;
        trav(u,i) {
            int v=e[i].v;
            if(dis[v]>dis[u]+e[i].w) {
                dis[v]=dis[u]+e[i].w;
                q.push((Node){v,dis[v]});
            }
        }
    }
}

  

  

最小生成树:

//Kruskal 

int fa[N];
int find(int u)
{
	if(!fa[u] || u==fa[u]) return fa[u]=u;
	return fa[u]=find(fa[u]);
}

struct Edge {
	int u,v,w;
	bool operator < (const Edge& rhs) const
	{
		return w<rhs.w;
	}
}e[M];
int tot;

void Kruskal()
{
	sort(e+1,e+tot+1);
	for(int i=1;i<=tot;i++) {
		int u=e[i].u,v=e[i].v;
		int x=find(u),y=find(v);
		if(x!=y) {
			fa[x]=y;
			//加入树边(u,v) 
		}
	}
}

  

最大流:

//Dinic算法求最大流 
struct Edge {
	int u,v,cap,flow;
};
struct Dinic {
	int d[N],cur[N],vis[N];
	vector<Edge> es;
	vector<int> g[N];
	queue<int> q;
	
	void AddEdge (int u,int v,int w) {
		es.push_back((Edge){u,v,w,0});
		es.push_back((Edge){v,u,0,0});
		int m=es.size();
		g[u].push_back(m-2);
		g[v].push_back(m-1);
	}
	bool bfs(int s,int t) {
		memset(vis,0,sizeof(vis));
		d[s]=0; vis[s]=1;
		q.push(s);
		while(!q.empty()) {
			int u=q.front(); q.pop();
			FOR(i,0,(int)g[u].size()-1) {
				Edge& e=es[g[u][i]];
				int v=e.v;
				if(e.cap>e.flow&&!vis[v]) {
					vis[v]=1;
					d[v]=d[u]+1;
					q.push(v);
				}
			}
		}
		return vis[t];
	}
	int dfs(int u,int a,int t) {
		if(u==t||a==0) return a;
		int flow=0,f;
		for(int& i=cur[u];i<g[u].size();i++) {
			Edge& e=es[g[u][i]];
			int v=e.v;
			if(d[v]==d[u]+1&&(f=dfs(v,min(a,e.cap-e.flow),t))>0) {
				e.flow+=f;
				es[g[u][i]^1].flow-=f;
				flow+=f,a-=f;
				if(!a) break;
			}
		}
		return flow;
	}
	int maxflow(int s,int t) {
		int flow=0;
		while(bfs(s,t)) {
			memset(cur,0,sizeof(cur));
			flow+=dfs(s,inf,t);
		}
		return flow;
	}
} dc;

  

最小费用最大流:

/最短路算法求最小费用最大流 
struct Edge {
	int u,v,cap,flow,cost;
	Edge(int _,int __,int ___,int ____,int _____)
	{ u=_,v=__,cap=___,flow=____,cost=_____; }
};

struct MCMF {
	int n,m,s,t;
	int d[N],p[N],a[N],inq[N];
	vector<Edge> es;
	vector<int> g[N];
	queue<int> q;
	void init(int n) {
		this->n=n;
		es.clear();
		for(int i=0;i<=n;i++) g[i].clear();
	}
	void AddEdge(int u,int v,int w,int c) {
		es.push_back(Edge(u,v,w,0,c));
		es.push_back(Edge(v,u,0,0,-c));
		int m=es.size();
		g[u].push_back(m-2);
		g[v].push_back(m-1);
	}
	bool spfa(int s,int t,ll& flow,ll& cost) {
		memset(inq,0,sizeof(inq));
		for(int i=0;i<=n;i++) d[i]=inf;
		inq[s]=1; d[s]=p[s]=0; a[s]=inf; 
		q.push(s);
		while(!q.empty()) {
			int u=q.front(); q.pop();
			inq[u]=0;
			for(int i=0;i<g[u].size();i++) {
				Edge& e=es[g[u][i]];
				int v=e.v;
				if(d[v]>d[u]+e.cost && e.cap>e.flow) {
					d[v]=d[u]+e.cost;
					a[v]=min(a[u],e.cap-e.flow);
					p[v]=g[u][i];
					if(!inq[v])
						inq[v]=1 , q.push(v);
				}
			}
		}
		if(d[t]==inf) return 0;
		flow+=a[t],cost+=a[t]*d[t];
		for(int x=t;x!=s;x=es[p[x]].u) {
			es[p[x]].flow+=a[t];
			es[p[x]^1].flow-=a[t];
		}
		return 1;
	}
	void mcmf(int s,int t,ll& cost,ll& flow) {
		flow=cost=0;
		while(spfa(s,t,cost,flow)) ;
	}
} mc;

 

KM算法:

//KM算法求二分图的最佳完美匹配 
struct KM {
	int slack[N],res[N];
	int l[N],r[N],lx[N],rx[N],g[N][N];
	
	void clear(int n) {
		for(int i=1;i<=n;i++) {
			res[i]=0;
			for(int j=1;j<=n;j++) g[i][j]=-1;
		}
	}
	bool find(int x,int n) {
		lx[x]=1;
		for(int i=1;i<=n;i++)
			if(!rx[i]&&g[x][i]!=-1) {
				int tmp=g[x][i]-l[x]-r[i];
				if(!tmp) {
					rx[i]=1;
					if(!res[i]||find(res[i],n)) {
						res[i]=x;
						return 1;
					}
				} else 
					slack[i]=min(slack[i],tmp);
			}
		return 0;
	}
	int solve(int n) {
		if(!n) return 0;
		for(int i=1;i<=n;i++) r[i]=0;
		for(int i=1;i<=n;i++) {
			l[i]=INF;
			for(int j=1;j<=n;j++) if(g[i][j]!=-1)
				l[i]=min(l[i],g[i][j]);
		}
		for(int i=1;i<=n;i++) {
			for(int j=1;j<=n;j++) slack[j]=INF;
			for(;;) {
				for(int j=1;j<=n;j++) lx[j]=rx[j]=0;
				if(find(i,n)) break;
				int mini=INF;
				for(int i=1;i<=n;i++) if(!rx[i])
					mini=min(mini,slack[i]);
				for(int i=1;i<=n;i++) {
					if(lx[i]) l[i]+=mini;
					if(rx[i]) r[i]-=mini;
					else slack[i]-=mini;
				}
			}
		}
		int ans=0;
		for(int i=1;i<=n;i++)
			ans+=l[i]+r[i];
		return ans;
	}
} km; 

  

 

 

LCA:

//倍增法求LCA
//倍增法可以在线构造 比较灵活 
struct Edge {
	int v,nxt;
}e[M];
int en=1,front[N];
void adde(int u,int v) 
{
	e[++en]=(Edge){v,front[u]}; front[u]=en;
}

int fa[N][D],dep[N];

void dfs(int u) 
{
	for(int i=1;i<D;i++)
		fa[u][i]=fa[fa[u][i-1]][i-1];
	trav(u,i) {
		int v=e[i].v;
		if(v!=fa[u][0]) {
			fa[v][0]=u;
			dep[v]=dep[u]+1;
			dfs(v);
		}
	}
}
int lca(int u,int v)
{
	if(dep[u]<dep[v]) swap(u,v);
	int t=dep[u]-dep[v];
	for(int i=0;i<D;i++)
		if(t&(1<<i)) u=fa[u][i];
	if(u==v) return u;
	for(int i=D-1;i>=0;i--)
		if(fa[u][i]!=fa[v][i])
			u=fa[u][i],v=fa[v][i];
	return fa[u][0];
}

//树链剖分求LCA
//比较快 
struct Edge {
	int v,nxt;
}e[M];
int en=1,front[N];
void adde(int u,int v)
{
	e[++en]=(Edge){v,front[u]}; front[u]=en;
}

int fa[N],top[N],siz[N],dep[N],son[N];
void dfs1(int u)
{
	siz[u]=1; son[u]=0;
	trav(u,i) {
		int v=e[i].v;
		if(v!=fa[u]) {
			fa[v]=u;
			dep[v]=dep[u]+1;
			dfs1(v);
			siz[u]+=siz[v];
			if(siz[v]>siz[son[u]]) son[u]=v;
		}
	}
}
void dfs2(int u,int tp)
{
	top[u]=tp; 
	if(son[u]) dfs2(son[u],tp);
	trav(u,i)
		if(e[i].v!=fa[u]&&e[i].v!=son[u])
			dfs2(e[i].v,e[i].v);
}
int lca(int u,int v)
{
	while(top[u]!=top[v]) {
		if(dep[top[u]]<dep[top[v]]) swap(u,v);
		u=fa[top[u]];
	}
	return dep[u]<dep[v]? u:v;
}

  

树链剖分:

//树链剖分
struct Edge {
	int v,nxt;
}e[M];
int en=1,front[N];
void adde(int u,int v)
{
	e[++en]=(Edge){v,front[u]}; front[u]=en;
}

int fa[N],top[N],siz[N],dep[N],son[N],bl[N],dfn;
void dfs1(int u)
{
	siz[u]=1; son[u]=0;
	trav(u,i) {
		int v=e[i].v;
		if(v!=fa[u]) {
			fa[v]=u;
			dep[v]=dep[u]+1;
			dfs1(v);
			siz[u]+=siz[v];
			if(siz[v]>siz[son[u]]) son[u]=v;
		}
	}
}
void dfs2(int u,int tp)
{
	top[u]=tp; bl[u]=++dfn;
	if(son[u]) dfs2(son[u],tp);
	trav(u,i)
		if(e[i].v!=fa[u]&&e[i].v!=son[u])
			dfs2(e[i].v,e[i].v);
}
//以合适的数据结构T维护重链 
int ans;
int query(int u,int v)
{
	while(top[u]!=top[v]) {
		if(dep[top[u]]<dep[top[v]]) swap(u,v);
		ans<-query(T,bl[top[u]],bl[u]);
		u=fa[top[u]];
	}
	if(u==v) return ;
	if(dep[u]>dep[v]) swap(u,v);
	ans<-query(T,bl[u],bl[v]);
	<-ans
}
//类似-查询树上任意两节点的方法 
void modify() {}

  

点分治:

//点分治 
struct Edge {
	int v,nxt;
}e[M];
int en=1,front[N];
void adde(int u,int v)
{
	e[++en]=(Edge){v,front[u]}; front[u]=en;
}

int rt,n,size,vis[N],siz[N],f[N],dep[N];

void get_root(int u,int fa)
{
	siz[u]=1; f[u]=0;
	trav(u,i) {
		int v=e[i].v;
		if(v!=fa) {
			get_root(v,u);
			siz[u]+=siz[v];
			if(siz[v]>f[u]) f[u]=siz[v];
		}
	}
	f[u]=max(f[u],size-siz[u]);
	if(f[u]<f[rt]) rt=u;
}
void solve(int u)
{
	vis[u]=1;
	//计算经过根u的信息 
	trav(u,i) if(!vis[e[i].v])
	{
		//统计当前子树信息
		//与前i-1个子树信息结合计算贡献
		//将当前子树信息加入前i-1个子树信息 
	}
	trav(u,i) if(!vis[e[i].v]) {
		int v=e[i].v;
		size=siz[v]; rt=0;
		get_root(v,-1);
		solve(rt); 
	}
}
int main()
{
	//blabla
	size=f[0]=n;
	rt=0; get_root(rt,-1);
	solve(rt);
}

  

 

字符串

 

KMP:

//KMP算法 
int f[N]; char s[N];
void get_fail()
{
    int j=0;
    int n=strlen(s+1);
    for(int i=2;i<=n;i++) {
        while(j&&s[j+1]!=s[i]) j=f[j];
        if(s[j+1]==s[i]) j++;
        f[i]=j;
    }
}

  

AC自动机:

//AC自动机 
struct AC_auto {
	int sz,ch[N][26],f[N],val[N];
	AC_auto() {
		sz=1;
		memset(ch,0,sizeof(ch));
	}
	void insert(char* s) {
		int u=0;
		for(int i=0;s[i];i++) {
			int c=s[i]-'a';
			if(!ch[u][c]) ch[u][c]=++sz;
			u=ch[u][c];
		}
		val[u]=1;
	}
	void get_fail() {
		queue<int> q;
		f[0]=0;
		for(int c=0;c<26;c++)
			if(ch[0][c]) f[ch[0][c]]=0,q.push(ch[0][c]);
		while(!q.empty()) {
			int qr=q.front(); q.pop();
			for(int c=0;c<26;c++) {
				int u=ch[qr][c];
				if(!u) continue;
				q.push(u);
				int v=f[qr];
				while(v&&!ch[v][c]) v=f[v];
				if(val[ch[v][c]]) val[u]=1;
				f[u]=ch[v][c];
			}
		}
	}
}; 

  

  

后缀自动机:

//后缀自动机SAM 
struct SAM {
	int sz,last,fa[N],ch[N][26],l[N];
	SAM() {
		sz=0; last=++sz;
		memset(l,0,sizeof(l));
		memset(fa,0,sizeof(fa));
	}
	void Add(int c) {
		int np=++sz,p=last; last=np;
		l[np]=l[p]+1;
		for(;p&&!ch[p][c];p=fa[p]) ch[p][c]=np;
		if(!p) fa[np]=1;
		else {
			int q=ch[p][c];
			if(l[q]==l[p]+1) fa[np]=q;
			else {
				int nq=++sz; l[nq]=l[p]+1;
				memcpy(ch[nq],ch[q],sizeof(ch[q]));
				fa[nq]=fa[q];
				fa[q]=fa[np]=nq;
				for(;q==ch[p][c];p=fa[p]) ch[p][c]=nq;
			}
		}
	}
	//do some other things
	
} sam;

  

后缀数组:

//后缀数组
#define rep(a,b,c) for(int a=(b);a>=(c);a--)
#define FOR(a,b,c) for(int a=(b);a<=(c);a++)
char s[N];
int c[N],t[N],t2[N],height[N],rank[N],sa[N];
void build_sa(int m,int n) {
	int *x=t,*y=t2,p,k;
	FOR(i,0,m-1) c[i]=0;
	FOR(i,0,n-1) c[x[i]=s[i]]++;
	FOR(i,0,m-1) c[i]+=c[i-1];
	rep(i,n-1,0) sa[--c[x[i]]]=i;
	
	for(k=1;k<=n;k<<=1) {
		p=0;
		FOR(i,n-k,n-1) y[p++]=i;
		FOR(i,0,n-1) if(sa[i]>=k) y[p++]=sa[i]-k;
		
		FOR(i,0,m-1) c[i]=0;
		FOR(i,0,n-1) c[x[y[i]]]++;
		FOR(i,0,m-1) c[i]+=c[i-1];
		rep(i,n-1,0) sa[--c[x[y[i]]]]=y[i];
		
		swap(x,y);
		p=1; x[sa[0]]=0;
		FOR(i,1,n-1)
			x[sa[i]]=y[sa[i]]==y[sa[i-1]]&&y[sa[i]+k]==y[sa[i-1]+k]? p-1:p++;
		if(p>=n) break;
		m=p;
	}
} 
void get_height(int n)
{
	FOR(i,0,n-1) rank[sa[i]]=i;
	int k=0;
	FOR(i,0,n-1) {
		if(k) k--;
		int j=sa[rank[i]-1];
		while(s[i+k]==s[j+k]) k++;
		height[rank[i]]=k;
	}
}

  

Manacher:

//Manacher算法 
char s[N],a[N];
int p[N];
void Add(int l,int r)
{
	l=l/2,r=r/2-1;
	if(l>r) return ;
	//q[++tot]=(Seg){l,r};
	//[l,r]为一个极大回文串 
}
void Manacher()
{
	int n=strlen(s+1);
	int m=n*2+1;
	for(int i=1;i<=n;i++) {
		a[i<<1]=s[i];
		a[i<<1|1]='#';
	}
	a[0]='+',a[1]='#',a[m+1]='-';
	int mx=0,id;
	for(int i=1;i<=m;i++) {
		if(mx>i) p[i]=min(mx-i,p[id*2-i]);
		else p[i]=1;
		while(a[i-p[i]]==a[i+p[i]]) p[i]++;
		Add(i-p[i],i+p[i]);
		if(p[i]+i>mx) mx=i+p[i],id=i;
	}
}

  

 

计算几何

 

计算几何基础知识:

//计算几何基础

const double eps = 1e-10;
int dcmp(double x) {
	if(fabs(x)<eps) return 0; else return x<0? -1:1;
}

struct Pt { 
	double x,y;
	Pt(double x=0,double y=0):x(x),y(y) {}
};
typedef Pt vec;

vec operator - (Pt A,Pt B) { return vec(A.x-B.x,A.y-B.y); }
vec operator + (vec A,vec B) { return vec(A.x+B.x,A.y+B.y); }
vec operator * (vec A,double p) { return vec(A.x*p , A.y*p); }
bool operator < (const Pt& a,const Pt& b) {
	return a.x<b.x || (a.x==b.x && a.y<b.y);
}
bool operator == (const Pt& a,const Pt& b) {
	return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
}

double cross(vec A,vec B) { return A.x*B.y-A.y*B.x; }
double Dot(vec A,vec B) { return A.x*B.x+A.y*B.y; }
double Len(vec A) { return sqrt(Dot(A,A)); }
double Angle(vec A,vec B) { return acos(Dot(A,B)/Len(A)/Len(B)); }

//逆时针旋转rad角度 
vec rotate(vec A,double rad) { 
	return vec(A.x*cos(rad)-A.y*sin(rad) , A.x*sin(rad)+A.y*cos(rad));
}
//法向量 左转90度 长度归1
vec Normal(vec A)
{
	double L=Len(A);
	return vec(-A.y/L,A.x/L);
}
//判断点在线段上 
bool OnSeg(Pt P,Pt a1,Pt a2) {
	return dcmp(cross(a1-P,a2-P))==0 && dcmp(Dot(a1-P,a2-P))<0;
}
//直线交点 
Pt LineIntersection(Pt P,vec v,Pt Q,vec w) {
	vec u=P-Q;
	double t=cross(w,u)/cross(v,w);
	return P+v*t;
}
double DistoLine(Pt P,Pt A,Pt B) {
	vec v1=B-A,v2=P-A;
	return fabs(cross(v1,v2))/Len(v1);
}
//线段不含端点 判断相交 
bool SegIntersection(Pt a1,Pt a2,Pt b1,Pt b2) {
	double c1=cross(a2-a1,b1-a1) , c2=cross(a2-a1,b2-a1) ,
		   c3=cross(b2-b1,a1-b1) , c4=cross(b2-b1,a2-b1);
	return dcmp(c1)*dcmp(c2)<0 && dcmp(c3)*dcmp(c4)<0;
	// b1 b2在线段a1a2的两侧  a1 a2在线段b1b2的两侧  规范相交
}
//线段含端点 判断线段严格相交 
bool SegInter(Pt s1, Pt e1, Pt s2, Pt e2) {
    if( min(s1.x, e1.x) <= max(s2.x, e2.x) &&
        min(s1.y, e1.y) <= max(s2.y, e2.y) &&
        min(s2.x, e2.x) <= max(s1.x, e1.x) &&
        min(s2.y, e2.y) <= max(s1.y, e1.y) &&
        cross(e1-s1,s2-s1) * cross(e1-s1,e2-s1) <= 0 &&
        cross(e2-s2,s1-s2) * cross(e2-s2,e1-s2) <= 0 
	  ) return true;
    return false;
}
//点到线段的距离 
double DistoSeg(Pt P,Pt A,Pt B) {
	if(A==B) return Len(P-A);
	vec v1=B-A , v2=P-A , v3=P-B;
	if(dcmp(Dot(v1,v2))<0) return Len(v2);
	else if(dcmp(Dot(v1,v3))>0) return Len(v3);
	else return fabs(cross(v1,v2))/Len(v1);
}
//多边形面积 
double PolygonArea(Pt* p,int n)
{
	double S=0;
	for(int i=1;i<n-1;i++)
		S+=cross(p[i]-p[0],p[i+1]-p[0]);
	return S/2;
}

  

凸包:

//凸包
const int N = 400000+10; 
const double PI = acos(-1.0);
const double eps = 1e-12;

int dcmp(double x) {
	if(fabs(x)<eps) return 0; else return x<0? -1:1;
}

struct Pt {
	double x,y;
	Pt(double x=0,double y=0) :x(x),y(y) {};
};
typedef Pt vec;

vec operator - (Pt a,Pt b) { return vec(a.x-b.x,a.y-b.y); }
vec operator + (vec a,vec b) { return vec(a.x+b.x,a.y+b.y); }
bool operator == (Pt a,Pt b) { 
	return dcmp(a.x-b.x)==0 && dcmp(a.y-b.y)==0;
}
bool operator < (const Pt& a,const Pt& b) {
	return a.x<b.x || (a.x==b.x && a.y<b.y);
}

vec rotate(vec a,double x) {
	return vec(a.x*cos(x)-a.y*sin(x),a.x*sin(x)+a.y*cos(x));
}
double cross(vec a,vec b) { return a.x*b.y-a.y*b.x; }
double dist(Pt a,Pt b) {
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

vector<Pt> ConvexHull(vector<Pt> p) {
	sort(p.begin(),p.end());
	p.erase(unique(p.begin(),p.end()),p.end());
	int n=p.size() , m=0;
	vector<Pt> ch(n+1);
	for(int i=0;i<n;i++) {
		while(m>1 && cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
		ch[m++]=p[i];
	}
	int k=m;
	for(int i=n-2;i>=0;i--) {
		while(m>k && cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0) m--;
		ch[m++]=p[i];
	}
	if(n>1) m--;
	ch.resize(m); return ch;
}

  

半平面交:

//半平面交

const int N =  305;
const double bond = 100001;
const double eps = 1e-10;

struct Pt {
	double x,y;
	Pt (double x=0,double y=0):x(x),y(y){}
};
typedef Pt vec;

vec operator + (Pt a,Pt b) { return vec(a.x+b.x,a.y+b.y); }
vec operator - (Pt a,Pt b) { return vec(a.x-b.x,a.y-b.y); }
vec operator * (Pt a,double p) { return vec(a.x*p,a.y*p); }

double cross(Pt a,Pt b) { return a.x*b.y-a.y*b.x; }

struct Line {
	Pt p; vec v; double ang;
	Line () {}
	Line (Pt p,vec v) :p(p),v(v){ ang=atan2(v.y,v.x); }
	bool operator < (const Line& rhs) const {
		return ang<rhs.ang;
	}
};

bool onleft(Line L,Pt p) { return cross(L.v,p-L.p)>0; }
Pt LineInter(Line a,Line b) 
{
	vec u=a.p-b.p;
	double t=cross(b.v,u)/cross(a.v,b.v);
	return a.p+a.v*t;
}
vector<Pt> HPI(vector<Line> L)
{
	int n=L.size();
	sort(L.begin(),L.end());
	int f,r;
	vector<Pt> p(n) , ans;
	vector<Line> q(n);
	q[f=r=0]=L[0];
	for(int i=1;i<n;i++) {
		while(f<r&&!onleft(L[i],p[r-1])) r--;
		while(f<r&&!onleft(L[i],p[f])) f++;
		q[++r]=L[i];
		if(fabs(cross(q[r].v,q[r-1].v))<eps) {
			r--;
			if(onleft(q[r],L[i].p)) q[r]=L[i];
		}
		if(f<r) p[r-1]=LineInter(q[r-1],q[r]);
	}
	while(f<r&&!onleft(q[f],p[r-1])) r--;
	if(r-f<=1) return ans;
	p[r]=LineInter(q[r],q[f]);
	for(int i=f;i<=r;i++) ans.push_back(p[i]);
	return ans; 
}

  

 

posted on 2016-04-05 11:21 hahalidaxin 阅读(...) 评论(...) 编辑 收藏

公告