Noip2016滚粗记QAQ

day1

t1

XBG

#include<map>
#include<cstdio>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int d[123456];
char ty[123456][20];
int n,m;
//... 
int main(){
    freopen("toy.in","r",stdin);
    freopen("toy.out","w",stdout);
    scanf("%d %d",&n,&m);
    for(int i=0;i<n;i++)scanf("%d%s",d+i,ty[i]);
    //for(int i=1;i<=n;i++)d[i]=d[i]?1:-1;
    int now=0;
    while(m--){
        int op,s;
        scanf("%d %d",&op,&s);
        if(!d[now]){
            if(op)now=(now+s)%n;
            else now=((now-s)%n+n)%n;    
        }else{
            if(op)now=((now-s)%n+n)%n;
            else now=(now+s)%n;
        }
    }
    cout<<ty[now];
    return 0;
}

t2

暴力30

子任务Si=1

以1为根统计每个子树里面1的个数

子任务链的写错了。

45分

#include<map>
#include<cstdio>
#include<string>
#include<vector>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
struct edge{
    int to,next;
}e[601234];
int cnt,last[301234],dep[301234],w[301234],ans[301234];
int n,m;
int s[301234],t[301234];
void insert(int a,int b){
    e[++cnt]=(edge){b,last[a]};last[a]=cnt;
    e[++cnt]=(edge){a,last[b]};last[b]=cnt;
}
int fa[301234][20];
void dfs(int v,int f){
    //fa[v][0]=f;
    //for(int i=1;i<=20;i++)fa[v][i]=fa[fa[v][i-1]][i-1];
    for(int i=last[v];i;i=e[i].next){
        int b=e[i].to;if(b!=f){
            dep[b]=dep[v]+1;
            dfs(b,v);
        }
    }
}
int an[301234];
void dfs2(int v,int t,int f){
    fa[v][0]=f;
    //for(int i=1;i<=20;i++)fa[v][i]=fa[fa[v][i-1]][i-1];
    for(int i=last[v];i;i=e[i].next){
        int b=e[i].to;if(b!=f){
            dfs2(b,t,v);
            if(b==t){
                an[v]=i;
            }
            if(an[b]){
                an[v]=i;
            }
        }
    }
}
void dfs3(int s,int t){
    int v=s,ti=0;
    ans[s]+=(ti==w[s]);
    while(s!=t){
        s=e[an[s]].to;
        ++ti;
        ans[s]+=(ti==w[s]);
    }
}
int siz[301234];
void dfs4(int v,int f){
    //fa[v][0]=f;
    //for(int i=1;i<=20;i++)fa[v][i]=fa[fa[v][i-1]][i-1];
    for(int i=last[v];i;i=e[i].next){
        int b=e[i].to;if(b!=f){
            dep[b]=dep[v]+1;
            dfs4(b,v);
            siz[v]+=siz[b];
            //fprintf(stderr,"Debug (%d)=%d\n",v,siz[v]);
        }
    }
}
vector<int>vec[301234];
int main(){
    freopen("running.in","r",stdin);
    freopen("running.out","w",stdout);
    scanf("%d %d",&n,&m);
    for(int i=1,u,v;i<n;i++){
        scanf("%d %d",&u,&v);
        insert(u,v);
    }
    //dfs(1,0);
    for(int i=1;i<=n;i++)scanf("%d",w+i);
    for(int i=1;i<=m;i++)scanf("%d %d",s+i,t+i);
    if(n<=1000){
        for(int i=1;i<=m;i++){
            memset(an,0,sizeof(an));
            dfs2(s[i],t[i],0);
            dfs3(s[i],t[i]);
        }
        for(int i=1;i<=n;i++)printf("%d ",ans[i]);
    }else
    if(n%10==4){
        for(int i=1;i<=n;i++)if(t[i]>=s[i])vec[i].push_back(s[i]);
        for(int i=1;i<=n;i++){
            if(i-w[i]>=0){
                int Ans=vec[i-w[i]].size(),cnt=0;
                for(int x=0;x<Ans;x++){
                    if(t[vec[i-w[i]][x]]<i)cnt++;
                }
                ans[i]+=(Ans-cnt);
            }
        }
        
        //for(int i=1;i<=n;i++)printf("%d ",ans[i]);
        for(int i=1;i<=n;i++)vec[i].clear();
        for(int i=1;i<=n;i++)if(t[i]<s[i])vec[i].push_back(s[i]);
        for(int i=1;i<=n;i++){
            if(i+w[i]<=n){
                int Ans=vec[i+w[i]].size(),cnt=0;
                //fprintf(stderr,"at %d\n",Ans);
                for(int x=0;x<Ans;x++){
                    if(t[vec[i+w[i]][x]]>i){
                        cnt++;
                    }
                }
                ans[i]+=(Ans-cnt);
            }
        }
        for(int i=1;i<=n;i++)printf("%d ",ans[i]);
    }else {
        for(int i=1;i<=n;i++)siz[t[i]]++;
        dep[1]=0; 
        dfs4(1,0);
        for(int i=1;i<=n;i++)printf("%d ",(w[i]==dep[i])*siz[i]);
    }
    return 0;//... //... //... 
}

t3

2^n*2^m暴力枚举所有申请和批准情况

 

 

暴力64,数组小了不然是72

要特判m<2

#include<map>
#include<cstdio>
#include<string>
#include<vector>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,v,e;
int c[333],d[333];
double k[333];/*
int min(int a,int b){
    return a<b?a:b;
}*/
struct zy{
    int i,cost;
}F[1234];
bool cmp(zy a,zy b){
    return a.cost>b.cost;
}
double brutef(){
    double ans=1e23;
    int _F=0;
    for(int i=1;i<=n;i++){
        int tmp=0;
        if(tmp>1)tmp=tmp+g[c[i-1]][c[i]]-g[c[i-1]][d[i]];
        if(tmp<n)tmp=tmp+g[c[i+1]][c[i]]-g[c[i+1]][d[i]];
        F[++_F]=(zy){i,-tmp};
    }
    sort(F+1,F+_F+1,cmp);
    int q[1234];
    memset(q,0,sizeof(q));
    for(int i=0;i<=m;i++){
        q[F[i].i]=1;double tmp2=0;
        for(int t=1;t<n;t++){
            int a=c[t],b=c[t+1];
            if(q[t])a=d[t];
            if(q[t+1])b=d[t+1];
            tmp2+=g[a][b];
        }
        //cout<<F[i].i<<","<<tmp2<<endl;
        ans=min(ans,tmp2);
    }
    return ans;
}
int main() {
    freopen("classroom.in","r",stdin);
    freopen("classroom.out","w",stdout);
    scanf("%d%d%d%d",&n,&m,&v,&e);
    for(int i=1;i<=n;i++)scanf("%d",c+i);
    for(int i=1;i<=n;i++)scanf("%d",d+i);
    bool xz2=1;
    for(int i=1;i<=n;i++)scanf("%lf",k+i),xz2=xz2&&(k[i]==1.0);
    
    for(int i=1;i<=v;i++)for(int j=1;j<=v;j++)g[i][j]=(i!=j)*(1<<20);
    for(int i=1;i<=e;i++){
        int a,b,w;
        scanf("%d %d %d",&a,&b,&w);
        g[a][b]=min(g[a][b],w);
        g[b][a]=min(g[b][a],w);
    }
    
    for(int p=1;p<=v;p++)for(int i=1;i<=v;i++)for(int j=1;j<=v;j++)g[i][j]=min(g[i][j],g[i][p]+g[p][j]);
    if(m<2){
        double ans=0,tmp=0;
        for(int i=1;i<n;i++){
            ans+=(g[c[i]][c[i+1]]);
        }
        tmp=ans;
        if(m)
        for(int x=1;x<=n;x++){
            //cout<<x<<" :\n";
            int res=0;
            for(int i=1;i<n;i++){
                int a=c[i],b=c[i+1];
                if(i==x)a=d[x];
                if(i+1==x)b=d[x];
                res+=g[a][b];
                
            }
            //cout<<res*k[x]+tmp*(1-k[x])<<endl;
            ans=min(ans,res*k[x]+tmp*(1-k[x]));
        }
        printf("%.2lf",(double)ans);
    }else if(n<=20&&m<=10){
        double ans=1e23;
        for(int i=0;i<(1<<n);i++){
            int p[30],q[33],cnt=0;
            for(int j=1;j<=n;j++)if(i&(1<<(j-1)))q[++cnt]=j;
            if(cnt<=m){
            //cerr<<"safe";
            memset(p,0,sizeof(p));
                double tmp=0;
                for(int j=0;j<(1<<cnt);j++){
                    double prob=1,tmp2=0;
                    for(int t=1;t<=cnt;t++){
                        if(j&(1<<(t-1))){
                            prob=prob*k[q[t]];
                            p[q[t]]=1;
                        }else {
                            prob=prob*(1-k[q[t]]);
                            p[q[t]]=0;
                        }
                    }
                    /*printf("(%d,%d)\n",i,j);
                    for(int t=1;t<=cnt;t++)cout<<q[t]<<"!!";cout<<endl;*/
                    for(int t=1;t<n;t++){
                        int a=c[t],b=c[t+1];
                        if(p[t])a=d[t];
                        if(p[t+1])b=d[t+1];
                        tmp2+=g[a][b];
                    }
                    //cout<<tmp2<<endl;;
                    tmp+=tmp2*prob;
                }
                //cout<<"---"<<tmp<<endl;
                ans=min(ans,tmp);
                }
            }//... 
        
        printf("%.2lf",(double)ans);
    }else{
        printf("%.2lf",(double)brutef());
    }
    
}

day2

t1

XBG

#include<cstdio>
#include<algorithm>
#include<iostream>
using namespace std;
int t,k,n,m,c[2333][2333],s[2333][2333];
int main(){
    freopen("problem.in","r",stdin);
    freopen("problem.out","w",stdout);
    scanf("%d%d",&t,&k);
    for(int i=0;i<=2000;i++)c[i][1]=i%k;
    for(int i=1;i<=2000;i++)
        for(int j=2;j<=i;j++)
            c[i][j]=(c[i-1][j]+c[i-1][j-1])%k;
    for(int i=1;i<=2000;i++)
        for(int j=1;j<=2000;j++){
            s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+(j<=i)*(!c[i][j]);
        }
    while(t--){
        scanf("%d %d",&n,&m);
        /*for(int i=1;i<=n;i++,puts(""))
            for(int j=1;j<=i;j++)printf("%d ",s[i][j]);
        */printf("%d\n",s[n][m]);
    }
    return 0;
}

t2

拿三个队列,一个存x,一个存px,一个存x-px。

然后这三个队列是单调递减的,具体证明较简单。

#include<queue>
#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
int n,m,q,u,v,t;double p;
int a[123456];
int PQ[123456],_PQ,od[7123456];
namespace solution{
    int x[7123456][2],y[7123456][2],z[7123456][2];
    int hx,hy,hz,tx,ty,tz;
    
    void init(){hx=hy=hz=1;tx=n;ty=tz=0;}
    void debug(int tim){
        for(int i=hx;i<=n;i++)printf("%d ",x[i][0]+q*(tim-x[i][1]));
        printf("!");for(int i=hy;i<=ty;i++)printf("%d ",y[i][0]+q*(tim-y[i][1]));
        printf("!");for(int i=hz;i<=tz;i++)printf("%d ",z[i][0]+q*(tim-z[i][1]));
        printf("!");
        puts("");
    }
    //wo hui zuo d2t2 ei
    void getres(){//debug(m);
        for(int i=1;i<=n+m;i++){
            int c=-12345;int *I;    
            if(hx<=n)if(x[hx][0]+q*(m-x[hx][1])>c)c=x[hx][0]+q*(m-x[hx][1]),I=&hx;
            if(hy<=ty)if(y[hy][0]+q*(m-y[hy][1])>c)c=y[hy][0]+q*(m-y[hy][1]),I=&hy;
            if(hz<=tz)if(z[hz][0]+q*(m-z[hz][1])>c)c=z[hz][0]+q*(m-z[hz][1]),I=&hz;
            ++*I;
            if(!(i%t))printf("%d ",c);
        }
    }
    int main(){
        sort(a+1,a+n+1,greater<int>());//nlogn
        for(int i=1;i<=n;i++)x[i][0]=a[i];
        p=u;p/=v;
        init();
        for(int i=1;i<=m;i++){
            int c=-12345;
            int *I;
            if(hx<=n)if(x[hx][0]+q*(i-1-x[hx][1])>c)c=x[hx][0]+q*(i-1-x[hx][1]),I=&hx;
            if(hy<=ty)if(y[hy][0]+q*(i-1-y[hy][1])>c)c=y[hy][0]+q*(i-1-y[hy][1]),I=&hy;
            if(hz<=tz)if(z[hz][0]+q*(i-1-z[hz][1])>c)c=z[hz][0]+q*(i-1-z[hz][1]),I=&hz;
            int A,B;
            A=p*c;
            B=c-A;
            //printf("cut %d into(%d,%d)\n",c,A,B);
            y[++ty][0]=A;
            z[++tz][0]=B;
            z[tz][1]=y[ty][1]=i;
            ++*I;
            od[i]=c;
            //cerr<<"h..="<<hx<<endl; 
            //debug(i);
        }
        for(int i=t;i<=m;i+=t)printf("%d ",od[i]);puts("");
        
        getres();
        return 0;
    }
}
int main(){
    freopen("earthworm.in","r",stdin);
    freopen("earthworm.out","w",stdout);
    //cerr<<sizeof(solution::x)*3;
    scanf("%d%d%d%d%d%d",&n,&m,&q,&u,&v,&t);
    for(int i=1;i<=n;i++)scanf("%d",a+i);
    if(n<=1000&&m<=1000){
        for(int i=1;i<=n;i++)PQ[i]=a[i];_PQ=n;
        for(int _t=1;_t<=m;_t++){
            int mx=1;
            for(int i=2;i<=_PQ;i++)if(PQ[i]>PQ[mx])mx=i;
            int x=PQ[mx];
            PQ[mx]=(int)u*1.0/v*x;
            PQ[++_PQ]=x-PQ[mx];
            //printf("cut %d into(%d,%d)\n",x,PQ[mx],PQ[_PQ]);
            for(int i=1;i<_PQ;i++)if(i!=mx)PQ[i]+=q;
            od[_t]=x;
        }
        for(int i=t;i<=m;i+=t)printf("%d ",od[i]);
        sort(PQ+1,PQ+n+m+1,greater<int>());
        puts("");
        for(int i=t;i<=n+m;i+=t)printf("%d ",PQ[i]);
    }else 
        solution::main();
    return 0;
}

t3

枚举两个点作抛物线,预处理能过什么点

状压(即或的值为2^n-1)

注意处理单点,分数未知

riio写错了

 

总结:d1t2 d2t3暴力写错了很伤

d2t2很像我之前写的QingdaoRegional的G

最后100+45+64+100+100+?未完待

posted @ 2016-11-20 15:46 zhouyis 阅读(...) 评论(...) 编辑 收藏