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
,tmpp
,t,tmpt,ans;
int d
,cur
,q
;
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\n";
#else
const char LL[]="%lld\n";
#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\n";
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\n";else cout<<"Yes\n";
}
}
return 0;
}