[CQOI2014]危桥 [网络流 最大流]

[BZOJ3504] [luoguP3163]

#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<cmath>
#include<stack>
using namespace std;
#define Min(x,y) ((x)<(y)?(x):(y))
const int N=100+50,M=40000+5,inf=0x3f3f3f3f,P=19650827;
int nw,n,a1,a2,an,b1,b2,bn,s,t,pre[N],incf[N];
char S[N][N];
template <class t>void rd(t &x){
    x=0;int w=0;char ch=0;
    while(!isdigit(ch)) w|=ch=='-',ch=getchar();
    while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    x=w?-x:x;
}

int head[N],tot=1;
struct edge{int v,flo,nxt;}e[N*N<<1];
void add(int u,int v,int flo){
    e[++tot]=(edge){v,flo,head[u]},head[u]=tot;
    e[++tot]=(edge){u,0,head[v]},head[v]=tot;
}

queue<int>q;bool vis[N];
bool bfs(){
    while(!q.empty()) q.pop();
    memset(vis,0,sizeof(vis));
    q.push(s),vis[s]=1,incf[s]=inf;
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=head[u],v;i;i=e[i].nxt)
        if(e[i].flo&&!vis[v=e[i].v]){
            incf[v]=Min(incf[u],e[i].flo);
            q.push(v),vis[v]=1,pre[v]=i;
            if(v==t) return 1;
        }
    }
    return 0;
}
void upd(){
    int x=t;
    while(x!=s){
        int i=pre[x];
        e[i].flo-=incf[t],e[i^1].flo+=incf[t];
        x=e[i^1].v;
    }
    nw+=incf[t];
}

int main(){
    freopen("in2.txt","r",stdin);
    //freopen("xor.out","w",stdout);
    while(scanf("%d%d%d%d%d%d%d",&n,&a1,&a2,&an,&b1,&b2,&bn)!=EOF){
        tot=1;memset(head,0,sizeof(head));
        for(int i=1;i<=n;++i){
            scanf("%s",S[i]+1);
            for(int j=1;j<=n;++j)
            if(i!=j){
                if(S[i][j]=='O') add(i-1,j-1,2);
                if(S[i][j]=='N') add(i-1,j-1,inf);
            }
        }
        s=n<<1|1,t=s+1;
        add(s,a1,an<<1),add(s,b1,bn<<1),add(a2,t,an<<1),add(b2,t,bn<<1);
        nw=0; while(bfs())
         upd();
        if(nw!=(an+bn)<<1) {puts("No");continue;}
        tot=1;memset(head,0,sizeof(head));
        for(int i=1;i<=n;++i)
            for(int j=1;j<=n;++j)
            if(i!=j){
                if(S[i][j]=='O') add(i-1,j-1,2);
                if(S[i][j]=='N') add(i-1,j-1,inf);
            }
        add(s,a1,an<<1),add(s,b2,bn<<1),add(a2,t,an<<1),add(b1,t,bn<<1);
        nw=0; while(bfs()) upd();
        if(nw!=(an+bn)<<1) puts("No");
        else puts("Yes");
    }
    return 0;
}

来源：https://www.cnblogs.com/lxyyyy/p/11415076.html