題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=4289
題意：現在，你就是國家安全局頭頭，有一羣猖狂的小偷，在城市S偷了無數的財富，現在他們要把這些財富運到城市D去，這個國家有N個城市（包括S和D），共有M條路聯通這N個城市。爲了阻止小偷將財富從S運到D，需要在每個城市建一個關卡，就需要一定的花費，問至少需要花費多少，能阻止小偷成功的將財富運到城市D。
思路：網絡流最小割。
代碼：
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
using namespace std;
#define inf 11000000
#define nMax 1005
#define mMax 100050
struct Node{
int u,v,next;
int c;
}edge[mMax];
int ne, head[nMax];
int cur[nMax], ps[nMax], dep[nMax],n,m,ans;
bool f[nMax],ff[nMax];
void addedge(int u, int v,int c){
edge[ne].u = u;
edge[ne].v = v;
edge[ne].c = c;
edge[ne].next = head[u];
head[u] = ne ++;
edge[ne].u = v;
edge[ne].v = u;
edge[ne].c = 0;
edge[ne].next = head[v];
head[v] = ne ++;
}
int dinic(int s, int t){
int tr, res = 0;
int i, j, k, f, r, top;
while(1){
memset(dep, -1, sizeof(dep));
for(f = dep[ps[0]=s] = 0, r = 1; f != r;)
for(i = ps[f ++], j = head[i]; j; j = edge[j].next)
if(edge[j].c && dep[k=edge[j].v] == -1){
dep[k] = dep[i] + 1;
ps[r ++] = k;
if(k == t){
f = r; break;
}
}
if(dep[t] == -1) break;
memcpy(cur, head, sizeof(cur));
i = s, top = 0;
while(1){
if(i == t){
for(tr = inf, k = 0; k < top; k ++)
if(edge[ps[k]].c < tr)
tr = edge[ps[f=k]].c;
for(k = 0; k < top; k ++){
edge[ps[k]].c -= tr;
edge[ps[k]^1].c += tr;
}
i = edge[ps[top=f]].u;
res += tr;
}
for(j = cur[i]; cur[i]; j = cur[i] = edge[cur[i]].next)
if(edge[j].c && dep[i]+1 == dep[edge[j].v]) break;
if(cur[i]){
ps[top ++] = cur[i];
i = edge[cur[i]].v;
}else{
if(top == 0) break;
dep[i] = -1;
i = edge[ps[-- top]].u;
}
}
}
return res;
}
int main(){
int i,a,b,N,M,ans;
int src,des;
int k;
while(scanf("%d%d",&N,&M)!=EOF)
{
scanf("%d%d",&src,&des);
ans=inf;
ne=2;
memset(head,0,sizeof(head));
for(i=1;i<=N;i++)
{
scanf("%d",&k);
if(i == des)
ans = k;
addedge(i,i+N,k);
}
while(M--)
{
scanf("%d%d",&a,&b);
addedge(a+N,b,inf);
addedge(b+N,a,inf);
}
int fuck=dinic(src,des);
if(fuck < ans) ans = fuck;
cout<<ans<<endl;
}
return 0;
}