BZOJ3504 CQOI2014危桥（最大流）

　　如果只有一个人的话很容易想到最大流，正常桥连限流inf双向边，危桥连限流2双向边即可。现在有两个人，容易想到给两起点建超源两汇点建超汇，但这样没法保证两个人各自到达自己要去的目的地。于是再超源连一个人的起点和另一个人的终点跑一遍，两次都满流说明有解。证明脑(bu)补(hui)。

#include<iostream> 
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
#define N 55
#define S 0
#define T 51
int n,s1,t1,c1,s2,t2,c2,p[N],tmpp[N],t,tmpt,ans;
int d[N],cur[N],q[N];
struct data{int to,nxt,cap,flow;
}edge[N*N<<2],tmpedge[N*N<<2];
void addedge(int x,int y,int z)
{
    t++;edge[t].to=y,edge[t].nxt=p[x],edge[t].cap=z,edge[t].flow=0,p[x]=t;
    t++;edge[t].to=x,edge[t].nxt=p[y],edge[t].cap=0,edge[t].flow=0,p[y]=t;
}
bool bfs()
{
    memset(d,255,sizeof(d));d[S]=0;
    int head=0,tail=1;q[1]=S;
    do
    {
        int x=q[++head];
        for (int i=p[x];~i;i=edge[i].nxt)
        if (d[edge[i].to]==-1&&edge[i].flow<edge[i].cap)
        {
            d[edge[i].to]=d[x]+1;
            q[++tail]=edge[i].to;
        }
    }while (head<tail);
    return ~d[T];
}
int work(int k,int f)
{
    if (k==T) return f;
    int used=0;
    for (int i=cur[k];~i;i=edge[i].nxt)
    if (d[k]+1==d[edge[i].to])
    {
        int w=work(edge[i].to,min(f-used,edge[i].cap-edge[i].flow));
        edge[i].flow+=w,edge[i^1].flow-=w;
        if (edge[i].flow<edge[i].cap) cur[k]=i;
        used+=w;if (used==f) return f;
    }
    if (used==0) d[k]=-1;
    return used;
}
void dinic()
{
    while (bfs())
    {
        memcpy(cur,p,sizeof(p));
        ans+=work(S,N<<2);
    }
}
int main()
{
#ifndef ONLINE_JUDGE
    freopen("bzoj3504.in","r",stdin);
    freopen("bzoj3504.out","w",stdout);
    const char LL[]="%I64d
";
#else
    const char LL[]="%lld
";
#endif
    while (scanf("%d%d%d%d%d%d%d",&n,&s1,&t1,&c1,&s2,&t2,&c2)>0)
    {
        s1++,t1++,s2++,t2++;c1<<=1,c2<<=1;
        memset(p,255,sizeof(p));t=-1;
        for (int i=1;i<=n;i++)
        {
            char c;
            for (int j=1;j<=n;j++)
            {
                c=getchar();
                while (c<'A'||c>'Z') c=getchar();
                if (c=='O') addedge(i,j,2);
                else if (c=='N') addedge(i,j,N<<2); 
            }
        }
        memcpy(tmpp,p,sizeof(p));
        memcpy(tmpedge,edge,sizeof(edge));
        tmpt=t;
        addedge(S,s1,c1);
        addedge(S,s2,c2);
        addedge(t1,T,c1);
        addedge(t2,T,c2);
        ans=0;
        dinic();
        if (ans<c1+c2) cout<<"No
";
        else
        {
            memcpy(p,tmpp,sizeof(p));
            memcpy(edge,tmpedge,sizeof(edge));
            t=tmpt;
            addedge(S,s1,c1);
            addedge(S,t2,c2);
            addedge(t1,T,c1);
            addedge(s2,T,c2);
            ans=0;
            dinic();
            if (ans<c1+c2) cout<<"No
";else cout<<"Yes
";
        }
    }
    return 0;
}