[CQOI2009]dance跳舞(最大流+二分)

[CQOI2009]dance跳舞

每个人拆成$2$个点,表示是否与喜欢的人跳舞

跳$m$首舞曲时,满足最大流为$n*m$

二分$m$,跑最大流即可

#include<cstdio>
#include<cstring>
inline int min(int A,int B){return A<B?A:B;}
#define W 9999
int n,k,S,T,cur[W],d[W],L,R,h[W]; 
char q[55][55]; bool vis[W];
int Cnt,hd[W],nxt[W],ed[W],poi[W],val[W];
void adde(int x,int y,int v){
    nxt[ed[x]]=++Cnt; hd[x]=hd[x]?hd[x]:Cnt;
    ed[x]=Cnt; poi[Cnt]=y; val[Cnt]=v;
}
inline void link(int x,int y,int v){adde(x,y,v),adde(y,x,0);}
#define to poi[i]
bool bfs(){
    for(int i=1;i<=T;++i) cur[i]=hd[i],vis[i]=0;
    h[L=1]=S; R=2; vis[S]=1;
    while(L!=R){
        int x=h[L++]; if(L>=W)L=1;
        for(int i=hd[x];i;i=nxt[i])
            if(!vis[to]&&val[i]){
                vis[to]=1,d[to]=d[x]+1;
                h[R++]=to; if(R>=W)R=1;
            }
    }return vis[T];
}
int dfs(int x,int a){
    if(x==T||!a) return a;
    int F=0,f;
    for(int &i=cur[x];i;i=nxt[i]){
        if(d[to]==d[x]+1&&(f=dfs(to,min(a,val[i])))>0)
            F+=f,a-=f,val[i]-=f,val[i^1]+=f;
        if(!a) break;
    }return F;
}
int dinic(){int re=0; while(bfs())re+=dfs(S,W); return re;}
int chk(int x){
    memset(hd,0,sizeof(hd)); Cnt=1;
    memset(ed,0,sizeof(ed));
    memset(nxt,0,sizeof(nxt));
    for(int i=1;i<=n;++i) link(S,i,x),link(i+n*3,T,x);
    for(int i=1;i<=n;++i){
        link(i,i+n,k); link(i+n*2,i+n*3,k);
        for(int j=1;j<=n;++j){
            if(q[i][j]=='Y') link(i,j+n*3,1);
            else link(i+n,j+n*2,1);
        }
    }return dinic()==x*n;
}
int main(){
    scanf("%d%d",&n,&k); S=n*4+1,T=S+1;
    for(int i=1;i<=n;++i) scanf("%s",q[i]+1);
    int l=1,r=n+1;
    while(l<r){
        int mid=(l+r)/2;
        if(chk(mid)) l=mid+1;
        else r=mid;
    }printf("%d",l-1);
    return 0;
}
原文地址:https://www.cnblogs.com/kafuuchino/p/11461532.html