- 题意: 最小割
- 思路: 点权为1,边权无限,要把点拆为出点和入点,由于指定了源点和汇点且源点汇点不能被割掉,超级源要连源点的出点,超级汇要连汇的入点。
#include<bits/stdc++.h>
#define ll long long
using namespace std;
typedef pair<int,int> pii;
const int N = 2000;
const int INF = 0x3f3f3f3f;
struct E{
int u,v,flow,nxt;
E(){}
E(int u,int v,int flow,int nxt):u(u),v(v),flow(flow),nxt(nxt){}
}e[N*2];
int n,m,sp,tp,tot;
int head[N],dis[N],pre[N],cur[N];
void init(){
tot = 0; memset(head,-1,sizeof head);
}
void addE(int u,int v,int flow){
e[tot].u = u; e[tot].v = v; e[tot].flow = flow; e[tot].nxt = head[u]; head[u] = tot++;
e[tot].u = v; e[tot].v = u; e[tot].flow = 0; e[tot].nxt = head[v]; head[v] = tot++;
}
int q[N];
int bfs(){
int qtop = 0,qend=0;
memset(dis,-1,sizeof dis);
dis[sp] = 0;
q[qend++] = sp;
while(qtop!=qend){
int u = q[qtop++];
if(u==tp) return true;
for(int i=head[u];~i;i=e[i].nxt){
int v = e[i].v;
if(dis[v]==-1 && e[i].flow){
dis[v] = dis[u]+1;
q[qend++] = v;
}
}
}
return dis[tp]!=-1;
}
int dfs(int u,int flow){
int res = 0;
if(u==tp) return flow;
for(int i=head[u];i!=-1&&flow;i=e[i].nxt){
int v = e[i].v;
if(dis[v]==dis[u]+1 && e[i].flow){
int d = dfs(v,min(e[i].flow,flow));
e[i].flow -=d;
e[i^1].flow += d;
res+=d;
flow -= d;
}
}
if(!res)
dis[u] = -2;
return res;
}
int dinic(){
int ans=0;
while(bfs()){
ans+=dfs(sp,INF);
}
return ans;
}
int main(){
int c1,c2;
scanf("%d%d%d%d",&n,&m,&c1,&c2);
sp = n*2 +1 , tp = sp +1;
int u,v;
init();
addE(sp,c1+n,INF);
addE(c2,tp,INF);
for(int i=1;i<=n;++i) addE(i,i+n,1);
for(int i=1;i<=m;++i){
scanf("%d%d",&u,&v);
addE(u+n,v,INF);
addE(v+n,u,INF);
}
printf("%d\n",dinic());
return 0;
}
洛谷 P1345 [USACO5.4]奶牛的电信Telecowmunication