2017-2018 ACM-ICPC Nordic Collegiate Programming Contest (NCPC 2017)

A. Airport Coffee

设$f_i$表示考虑前$i$个咖啡厅,且在$i$处买咖啡的最小时间,通过单调队列优化转移。

时间复杂度$O(n)$。

#include<cstdio>
const int N=500010;
typedef long long ll;
const double inf=1e20;
ll L,A,B,T,R,lim0,lim1,a[N];
double f[N],val[N];
double dp0=inf;
int id0;
int pre[N];
int i,j,k;
int h=1,t,q[N];
double ans;
int fin;
int cof[N],cnt,n;
int main(){
	scanf("%lld%lld%lld%lld%lld",&L,&A,&B,&T,&R);
	lim0=A*T+R*B;
	lim1=T*A;
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%lld",&a[i]);
	ans=1.0*L/A;
	for(i=j=k=1;i<=n;i++){
		f[i]=1.0*a[i]/A;
		
		while(j<i&&a[i]-a[j]>=lim0){
			double now=f[j]-1.0*a[j]/A;
			if(now<dp0){
				dp0=now;
				id0=j;
			}
			j++;
		}
		if(id0){
			double now=dp0+1.0*(a[i]-lim0)/A+T+R;
			if(now<f[i]){
				f[i]=now;
				pre[i]=id0;
			}
		}
		
		while(k<i&&a[i]-a[k]>=lim1){
			val[k]=f[k]-1.0*a[k]/B;
			while(h<=t&&val[q[t]]>val[k])t--;
			q[++t]=k;
			k++;
		}
		while(h<=t&&a[i]-a[q[h]]>=lim0)h++;
		if(h<=t){
			double now=val[q[h]]+1.0*(a[i]-lim1)/B+T;
			if(now<f[i]){
				f[i]=now;
				pre[i]=q[h];
			}
		}
		double ret=f[i];
		ll d=L-a[i];
		if(d<=lim1){
			ret+=1.0*d/A;
		}else if(d<=lim0){
			ret+=1.0*(d-lim1)/B+T;
		}else{
			ret+=1.0*(d-lim0)/A+T+R;
		}
		if(ret<ans){
			ans=ret;
			fin=i;
		}
	}
	while(fin){
		cof[++cnt]=fin;
		fin=pre[fin];
	}
	printf("%d\n",cnt);
	for(i=cnt;i;i--)printf("%d ",cof[i]-1);
}
//0 .. n-1

  

B. Best Relay Team

按题意模拟即可。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() {  }
#define MS(x, y) memset(x, y, sizeof(x))
#define MC(x, y) memcpy(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1010, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int l, r;
struct A
{
	string s;
	double fi, se;
}a[N], b[N];
int n;
string ss[5];

bool cmp(A a, A b)
{
	return a.se < b.se;
}
int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; i ++){
		cin >> a[i].s;
		scanf("%lf%lf", &a[i].fi, &a[i].se);
		a[i + n] = a[i];
	}
	double ans = 1e9;
	for(int i = 1; i <= n; i ++){
		MC(b, a);
		sort(b + i + 1, b + n + i, cmp);
		double tmp = b[i].fi + b[i + 1].se + b[i + 2].se + b[i + 3].se; 
		if(tmp < ans){
			ans = tmp;
			ss[1] = b[i].s;
			ss[2] = b[i + 1].s;
			ss[3] = b[i + 2].s;
			ss[4] = b[i + 3].s;
		}
	}
	printf("%.2f\n", ans);
	for(int i = 1; i <= 4; i ++){
		//printf("%s\n", ss[i]);	
		cout << ss[i] << endl;
	}
	scanf("%d", &n);
	
	return 0;
}

/*
【trick&&吐槽】


【题意】


【分析】


【时间复杂度&&优化】


*/

  

C. Compass Card Sales

按题意模拟即可。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() {  }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n;
struct A
{
	int v[3];
	int id;
	int val;
}a[N];// keep a not modified
struct B
{
	int id;
	int val;
	bool operator < (const B & b)const
	{
		if(val != b.val)return val < b.val;
		return id > b.id;
	}
};
set<int>sot[3][720];
set<B>b;
//use the a[o]
int getval(int o)
{
	int val = 0;
	for(int i = 0; i < 3; ++i)
	{
		int pos = a[o].v[i];
		if(sot[i][pos].size() >= 2)
		{
			continue;
		}
		int pre = pos + 360;
		do
		{		
			--pre;
		}while(!sot[i][pre].size());
		int suf = pos;
		do
		{
			++suf;
		}while(!sot[i][suf].size());
		val += (pos + 360 - pre) + (suf - pos);
	}
	//printf("o = %d a[o].id = %d a[o].val = %d\n", o, a[o].id, val);
	return val;
}
map<int, int>mop;
int main()
{
	while(~scanf("%d",&n))
	{
		mop.clear();
		for(int i = 0; i < 3; ++i)
		{
			for(int j = 0; j < 720; ++j)
			{
				sot[i][j].clear();
			}
		}
		b.clear();
		for(int i = 1; i <= n; ++i)
		{
			for(int j = 0; j < 3; ++j)
			{
				scanf("%d", &a[i].v[j]);
			}
			scanf("%d", &a[i].id);
			mop[a[i].id] = i;
			for(int j = 0; j < 3; ++j)
			{
				int p = a[i].v[j];
				int o = a[i].id;
				sot[j][p].insert(o);
				sot[j][p + 360].insert(o);
			}
		}
		//a index is a input order
		for(int i = 1; i <= n; ++i)
		{
			a[i].val = getval(i);
			b.insert({a[i].id, a[i].val});
		}
		while(n--)
		{
			int id = b.begin()->id;
			int o = mop[id];
			printf("%d\n", id);
			b.erase({id, a[o].val});
			for(int i = 0; i < 3; ++i)
			{
				int pos = a[o].v[i];
				sot[i][pos].erase(id);
				sot[i][pos + 360].erase(id);
			}
			
			for(int i = 0; i < 3; ++i)
			{
				int pos = a[o].v[i];
				if(sot[i][pos].size() == 1)
				{
					int rst = mop[*sot[i][pos].begin()];
					b.erase({a[rst].id, a[rst].val});
					a[rst].val = getval(rst);
					b.insert({a[rst].id, a[rst].val});
				}
				else if(sot[i][pos].size() == 0 && n > 1)
				{		
					int pos = a[o].v[i];
					int pre = pos + 360;
					do
					{		
						--pre;
					}while(!sot[i][pre].size());
					if(sot[i][pre].size() == 1)
					{
						int rst = mop[*sot[i][pre].begin()];
						b.erase({a[rst].id, a[rst].val});
						a[rst].val = getval(rst);
						b.insert({a[rst].id, a[rst].val});
					}
					int suf = pos;
					do
					{
						++suf;
					}while(!sot[i][suf].size());
					if(sot[i][suf].size() == 1)
					{
						int rst = mop[*sot[i][suf].begin()];
						b.erase({a[rst].id, a[rst].val});
						a[rst].val = getval(rst);
						b.insert({a[rst].id, a[rst].val});
					}
				}
			}
		}
	}
	return 0;
}

/*
【trick&&吐槽】


【题意】


【分析】


【时间复杂度&&优化】


*/

  

D. Distinctive Character

设$f_S$表示到$S$的最大相似度,显然对于输入的$n$个串$a_i$,有$f_{a_i}=k$。

对于$f_S=k$,将$S$修改一位,可以得到$f_{S'}=k-1$,BFS求出所有$S$即可。

时间复杂度$O(k2^k)$。

#include<cstdio>
int n,m,i,h,t,f[1<<20],q[2222222],x,y;
char s[100];
inline void ext(int x,int y){if(f[x]<0)f[q[++t]=x]=y;}
int main(){
	scanf("%d%d",&n,&m);
	for(i=0;i<1<<m;i++)f[i]=-1;
	while(n--){
		scanf("%s",s);
		int t=0;
		for(i=0;i<m;i++)t=t*2+s[i]-'0';
		f[t]=m;
	}
	h=1;
	for(i=0;i<1<<m;i++)if(~f[i])q[++t]=i;
	while(h<=t){
		x=q[h++];
		y=f[x]-1;
		for(i=0;i<m;i++)ext(x^(1<<i),y);
	}
	for(i=m-1;~i;i--)printf("%d",x>>i&1);
}

  

E. Emptying the Baltic

令两点间的边权为较大的高度,求出最小生成树,则每个点的实际水位为起点到它路径上点权的最大值。

#include<cstdio>
#include<algorithm>
using namespace std;
const int N=550,M=N*N;
int n,m,i,j,x,y,tot,id[N][N],a[M],f[M];
int ce;
int v[M*2],nxt[M*2],ed,g[M];
long long ans;
struct E{int x,y,z;E(){}E(int _x,int _y,int _z){x=_x,y=_y,z=_z;}}e[M*9];
inline bool cmp(const E&a,const E&b){return a.z<b.z;}
int F(int x){return f[x]==x?x:f[x]=F(f[x]);}
inline void add(int x,int y){
	v[++ed]=y;nxt[ed]=g[x];g[x]=ed;
}
void dfs(int x,int y,int z){
	z=max(z,a[x]);
	if(z<0)ans-=z;
	for(int i=g[x];i;i=nxt[i])if(v[i]!=y)dfs(v[i],x,z);
}
int main(){
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++)for(j=1;j<=m;j++){
		id[i][j]=++tot;
		scanf("%d",&a[tot]);
	}
	for(i=1;i<=n;i++)for(j=1;j<=m;j++)for(x=-1;x<=1;x++)for(y=-1;y<=1;y++){
		int nx=i+x,ny=j+y;
		if(nx<1||nx>n||ny<1||ny>m)continue;
		e[++ce]=E(id[i][j],id[nx][ny],max(a[id[i][j]],a[id[nx][ny]]));
	}
	sort(e+1,e+ce+1,cmp);
	for(i=1;i<=tot;i++)f[i]=i;
	for(i=1;i<=ce;i++)if(F(e[i].x)!=F(e[i].y)){
		f[f[e[i].x]]=f[e[i].y];
		add(e[i].x,e[i].y);
		add(e[i].y,e[i].x);
	}
	scanf("%d%d",&i,&j);
	dfs(id[i][j],0,-10000000);
	printf("%lld",ans);
}

  

F. Fractal Tree

留坑。

 

G. Galactic Collegiate Programming Contest

用一个set维护所有严格比$1$好的队伍,若别人过题则试情况插入set,若$1$过题则将set中最差的若干队伍删除。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() {  }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, m;
struct A
{
	int g, t;
	bool operator < (const A &b)const
	{
		if(g != b.g)return g > b.g;
		return t < b.t;
	}
}a[N];
multiset<A>sot;
int main()
{
	while(~scanf("%d%d",&n, &m))
	{
		for(int i = 1; i <= n; ++i)
		{
			a[i].g = a[i].t = 0;
		}
		for(int i = 1; i <= m; ++i)
		{
			int o, p;
			scanf("%d%d", &o, &p);
			if(o == 1)
			{
				a[1].g += 1;
				a[1].t += p;
			}
			else
			{
				if(a[o] < a[1])
				{
					sot.erase(sot.find(a[o]));
				}
				a[o].g += 1;
				a[o].t += p;
				sot.insert(a[o]);
			}
			while(!sot.empty() && !(*--sot.end() < a[1]))
			{
				sot.erase(--sot.end());
			}
			printf("%d\n", sot.size() + 1);
		}
	}
	return 0;
}

/*
【trick&&吐槽】


【题意】


【分析】


【时间复杂度&&优化】


*/

  

H. Hubtown

极角排序后求出每个人最近的$1$到$2$条火车线路,然后求最大流即可。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() {  }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long ll;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 4e5 + 10, M = 2e6 + 10, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, m;
int ST, ED, ID;
int first[N], w[M], cap[M], nxt[M];
void ins(int x, int y, int cap_)
{
	w[++ID] = y;
	cap[ID] = cap_;
	nxt[ID] = first[x];
	first[x] = ID;
	w[++ID] = x;
	cap[ID] = 0;
	nxt[ID] = first[y];
	first[y] = ID;
}
int d[N];
bool bfs()
{
	MS(d, -1); d[ST] = 0;
	queue<int>q; q.push(ST);
	while(!q.empty())
	{
		int x = q.front(); q.pop();
		for(int z = first[x]; z; z = nxt[z])if(cap[z])
		{
			int y = w[z];
			if(d[y] == -1)
			{
				d[y] = d[x] + 1;
				if(y == ED)return 1;
				q.push(y);
			}
		}
	}
	return 0;
}
int dfs(int x, int all)
{
	if(x == ED)return all;
	int use = 0;
	for(int z = first[x]; z; z = nxt[z])if(cap[z])
	{
		int y = w[z];
		if(d[y] == d[x] + 1)
		{
			int tmp = dfs(y, min(cap[z], all - use));
			cap[z] -= tmp;
			cap[z ^ 1] += tmp;
			use += tmp;
			if(use == all)break;
		}
	}
	if(use == 0)d[x] = -1;
	return use;
}
int dinic()
{
	int tmp = 0;
	while(bfs())tmp += dfs(ST, inf);
	return tmp;
}
struct A
{
	int x, y, o;
}a[N];
struct B
{
	int x, y, o,c;
}b[N];
struct E{
	int x,y,t;
	int sgn;
	E(){}
	E(int _x,int _y,int _t){
		x=_x,y=_y,t=_t;
		if(!x)sgn=y>0;
		else sgn=x>0;
	}
}e[400010];
inline bool cmpe(const E&a,const E&b){
	if(a.sgn!=b.sgn)return a.sgn<b.sgn;
	int t=a.x*b.y-a.y*b.x;
	if(t)return t<0;
	return a.t<b.t;
}
int pre[400010],suf[400010];
int gcd(int a,int b){return b?gcd(b,a%b):a;}
inline bool point_on_line(const A&a,const B&b){
	int d1=gcd(abs(a.x),abs(a.y)),d2=gcd(abs(b.x),abs(b.y));
	return (a.x/d1==b.x/d2)&&(a.y/d1==b.y/d2);
}
struct Point{
	ll x,y;
	Point(){}
	Point(ll _x,ll _y){x=_x,y=_y;}
};
inline ll cross(const Point&a,const Point&b){return a.x*b.y-a.y*b.x;}
inline ll sig(ll x){
	if(x==0)return 0;
	return x>0?1:-1;
}
typedef long double ld;


const ld epsss=1e-10;

struct Pointd{
	ld x,y;
	Pointd(){}
	Pointd(ld _x,ld _y){x=_x,y=_y;}
};
inline ld crossd(const Pointd&a,const Pointd&b){return a.x*b.y-a.y*b.x;}
inline ll sigd(ld x){
	if(fabs(x)<epsss)return 0;
	return x>0?1:-1;
}

inline int distance_cmp(const A&_a,const B&_b,const B&_c){
	Point a(_a.x,_a.y);
	Point b(_b.x,_b.y);
	Point c(_c.x,_c.y);
	Point d;
	if(!cross(b,c)){
		d=Point(-b.y,b.x);
		if(!cross(a,d))return 0;
		if(sig(cross(d,a))==sig(cross(d,b)))return -1;
		return 1;
	}
	ld L=sqrt(b.x*b.x+b.y*b.y);
	ld R=sqrt(c.x*c.x+c.y*c.y);
	Pointd aa(a.x,a.y);
	Pointd bb(b.x,b.y);
	Pointd cc(c.x,c.y);
	Pointd dd(d.x,d.y);
	bb.x*=R;
	bb.y*=R;
	cc.x*=L;
	cc.y*=L;
	dd=Pointd(bb.x+cc.x,bb.y+cc.y);
	if(!sigd(crossd(aa,dd)))return 0;
	if(sigd(crossd(dd,aa))==sigd(crossd(dd,bb)))return -1;
	return 1;
}
int main()
{
	while(~scanf("%d%d",&n, &m))
	{
		ST = 0;
		ED = n + m + 1;
		ID = 1; MS(first, 0);
		for(int i = 1; i <= n; ++i)
		{
			scanf("%d%d", &a[i].x, &a[i].y);
			a[i].o = i;
		}
		//sort a?
		
		for(int i = 1; i <= m; ++i)
		{
			scanf("%d%d%d", &b[i].x, &b[i].y,&b[i].c);
			b[i].o = i;
		}
		//sort b?
		int ce=0;
		for(int i=1;i<=n;i++){
			e[++ce]=E(a[i].x,a[i].y,i);
		}
		for(int i=1;i<=m;i++){
			e[++ce]=E(b[i].x,b[i].y,-i);
		}
		sort(e+1,e+ce+1,cmpe);
		
		pre[0]=0;
		for(int i=1;i<=ce;i++)if(e[i].t<0)pre[0]=-e[i].t;
		for(int i=1;i<=ce;i++){
			pre[i]=pre[i-1];
			if(e[i].t<0)pre[i]=-e[i].t;
		}
		
		suf[ce+1]=0;
		for(int i=ce;i;i--)if(e[i].t<0)suf[ce+1]=-e[i].t;
		for(int i=ce;i;i--){
			suf[i]=suf[i+1];
			if(e[i].t<0)suf[i]=-e[i].t;
		}
		
		for(int i=1;i<=ce;i++)if(e[i].t>0){
			int x=e[i].t;
			int L=pre[i],R=suf[i];
			if(L==R){
				ins(x,L+n,1);
				continue;
			}
			if(point_on_line(a[x],b[L])){
				ins(x,L+n,1);
				continue;
			}
			if(point_on_line(a[x],b[R])){
				ins(x,R+n,1);
				continue;
			}
			int t=distance_cmp(a[x],b[L],b[R]);
			if(t<=0)ins(x,L+n,1);
			if(t>=0)ins(x,R+n,1);
		}
		for(int i = 1; i <= n; ++i)
		{
			ins(ST, i, 1);
		}
		for(int i = 1; i <= m; ++i)
		{
			ins(n + i, ED, b[i].c);
		}
		
		//addedge: &&maybe use a little greedy method
		
		printf("%d\n",dinic());
		for(int i = 1; i <= n; ++i)
		{
			for(int z = first[i]; z; z = nxt[z])
			{
				int y = w[z];
				if(y > n && cap[z] == 0)
				{
					printf("%d %d\n", i - 1, y - n - 1);
				}
			}
		}
	}
	return 0;
}

/*
【trick&&吐槽】


【题意】


【分析】


【时间复杂度&&优化】


*/

  

I. Import Spaghetti

枚举起点,BFS求最小环。

#include<cstdio>
#include<map>
#include<cstring>
#include<iostream>
#include<sstream>
using namespace std;
const int N=555;
int n,i;
string name[N];
bool g[N][N];
map<string,int>id;
int ans=N,finS,finT;
int d[N],f[N],q[N],h,t;
inline int ask(string t){
	string x="";
	for(int i=0;i<t.size();i++)if(t[i]>='a'&&t[i]<='z')x.push_back(t[i]);
	return id[x];
}
inline void ext(int x,int y,int z){
	if(d[x]<0){
		q[++t]=x;
		d[x]=y;
		f[x]=z;
	}
}
void bfs(int S){
	int i,j,x;
	for(i=1;i<=n;i++)d[i]=-1;
	h=1,t=0;
	ext(S,0,0);
	while(h<=t){
		x=q[h++];
		int y=d[x]+1;
		for(i=1;i<=n;i++)if(g[x][i])ext(i,y,x);
	}
	for(i=1;i<=n;i++)if(d[i]>0&&g[i][S]){
		int now=d[i]+1;
		if(now<ans)ans=now,finS=S,finT=i;
	}
}
int main(){
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		cin>>name[i];
		id[name[i]]=i;
	}
	for(i=1;i<=n;i++){
		string tmp;int k;
		cin>>tmp>>k;
		char ss[2]; gets(ss);
		while(k--)
		{
			string s;
			getline(cin, s);
			stringstream cinn(s);
			int flag=1;
			while(cinn>>s)
			{
				if(flag){flag=0;continue;}
				int y=ask(s);
				g[y][i]=1;
				//printf("%d %d\n",i,y);
			}
		}
	}
	for(i=1;i<=n;i++)if(g[i][i]){
		cout<<name[i]<<endl;
		return 0;
	}
	for(i=1;i<=n;i++)bfs(i);
	if(ans>=N){
		puts("SHIP IT");
		return 0;
	}
	//printf("%d %d %d\n",ans,finS,finT);
	bfs(finS);
	i=finT;
	while(i){
		cout<<name[i]<<" ";
		i=f[i];
	}
}
/*
4
a b c d
a 1
import d, b, c
b 2
import d
import c
c 1
import c
d 0


5
classa classb myfilec execd libe
classa 2
import classb
import myfilec, libe
classb 1
import execd
myfilec 1
import libe
execd 1
import libe
libe 0


5
classa classb myfilec execd libe
classa 2
import classb
import myfilec, libe
classb 1
import execd
myfilec 1
import libe
execd 1
import libe, classa
libe 0
*/

  

J. Judging Moose

按题意模拟即可。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() {  }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int l, r;
int main()
{
	while(~scanf("%d%d",&l, &r))
	{
		if(l + r == 0)
		{
			puts("Not a moose");
		}
		else if(l == r)
		{
			printf("Even %d\n", l + r);
		}
		else 
		{
			printf("Odd %d\n", max(l, r) * 2);
		}
	}
	return 0;
}

/*
【trick&&吐槽】


【题意】


【分析】


【时间复杂度&&优化】


*/

  

K. Kayaking Trip

二分答案,贪心配对检验。

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() {  }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 1e5 + 10, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int n, m;
int g[3], gg[3];
int s[3];
int c[N];
bool check(int aim)
{
	gg[0] = g[0];
	gg[1] = g[1];
	gg[2] = g[2];
	for(int i = 1; i <= m; ++i)
	{
		int need = aim / c[i] + (aim % c[i] > 0);
		int jj = -1;
		int kk = -1;
		for(int j = 0; j < 3; ++j)
		{
			for(int k = j; k < 3; ++k)
			{
				int num = 1;
				if(j == k)num = 2;
				if(gg[j] >= num && gg[k] >= num && s[j] + s[k] >= need)
				{					
					if(jj == -1 || s[j] + s[k] < s[jj] + s[kk])
					{
						jj = j;
						kk = k;
					}
				}
			}
		}
		if(jj == -1)return 0;
		gg[jj] -= 1;
		gg[kk] -= 1;
		//
		//printf("after ope %d: %d %d %d\n", i, gg[0], gg[1], gg[2]);
		//
	}
	return 1;
}
int main()
{
	while(~scanf("%d%d%d",&g[0], &g[1], &g[2]))
	{
		scanf("%d%d%d",&s[0], &s[1], &s[2]);
		m = (g[0] + g[1] + g[2]) / 2;
		for(int i = 1; i <= m; ++i)scanf("%d", &c[i]);
		sort(c + 1, c + m + 1);
		int l = c[1] * (s[0] + s[0]);
		int r = c[m] * (s[2] + s[2]);
		//
		//l = r = 505;
		//printf("l == %d r == %d\n", l, r);
		//
		int ans = -1;
		while(l <= r)
		{
			int mid = (l + r + 1) / 2;
			if(check(mid))
			{
				ans = mid;
				l = mid + 1;
			}
			else
			{
				r = mid - 1;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}

/*
【trick&&吐槽】


【题意】


【分析】


【时间复杂度&&优化】


*/

  

posted @ 2017-11-30 00:02 Claris 阅读(...) 评论(...) 编辑 收藏