Control

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 238    Accepted Submission(s): 135

Problem Description 　　You, the head of Department of Security, recently received a top-secret information that a group of terrorists is planning to transport some WMD
¹ from one city (the source) to another one (the destination). You know their date, source and destination, and they are using the highway network.

　　The highway network consists of bidirectional highways, connecting two distinct city. A vehicle can only enter/exit the highway network at cities only.

　　You may locate some SA (special agents) in some selected cities, so that when the terrorists enter a city under observation (that is, SA is in this city), they would be caught immediately.

　　It is possible to locate SA in all cities, but since controlling a city with SA may cost your department a certain amount of money, which might vary from city to city, and your budget might not be able to bear the full cost of controlling all cities, you must identify a set of cities, that:

　　* all traffic of the terrorists must pass at least one city of the set.

　　* sum of cost of controlling all cities in the set is minimal.

　　You may assume that it is always possible to get from source of the terrorists to their destination.

————————————————————

¹ Weapon of Mass Destruction  

Input 　　There are several test cases.

　　The first line of a single test case contains two integer N and M ( 2 <= N <= 200; 1 <= M <= 20000), the number of cities and the number of highways. Cities are numbered from 1 to N.

　　The second line contains two integer S,D ( 1 <= S,D <= N), the number of the source and the number of the destination.

　　The following N lines contains costs. Of these lines the ith one contains exactly one integer, the cost of locating SA in the ith city to put it under observation. You may assume that the cost is positive and not exceeding 10
⁷.

　　The followingM lines tells you about highway network. Each of these lines contains two integers A and B, indicating a bidirectional highway between A and B.

　　Please process until EOF (End Of File).  

Output 　　For each test case you should output exactly one line, containing one integer, the sum of cost of your selected set.

　　See samples for detailed information.  

Sample Input 5 6 5 3 5 2 3 4 12 1 5 5 4 2 3 2 4 4 3 2 1  

Sample Output 3  

Source
2012 ACM/ICPC Asia Regional Chengdu Online  

Recommend liuyiding     成都赛区网络赛1002题。 一场网络赛两题最大流，用SAP模板可以秒掉。

//1002
/*
HDU 4289
G++  62ms  1888K
最大流
SAP
*/
#include<stdio.h>
#include<iostream>
#include<map>
#include<set>
#include<algorithm>
#include<string.h>
#include<stdlib.h>
using namespace std;

const int MAXN=5000;//点数的最大值
const int MAXM=2500000;//边数的最大值
const int INF=0x3f3f3f3f;

struct Node
{
    int from,to,next;
    int cap;
}edge[MAXM];
int tol;
int head[MAXN];
int dep[MAXN];
int gap[MAXN];//gap[x]=y:说明残留网络中 dep[i]==x的个数为y

int n;//点的实际个数，一定是总的点的个数，包括源点和汇点
void init()
{
    tol=0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w)
{
    edge[tol].from=u;
    edge[tol].to=v;
    edge[tol].cap=w;
    edge[tol].next=head[u];
    head[u]=tol++;
    edge[tol].from=v;
    edge[tol].to=u;
    edge[tol].cap=0;
    edge[tol].next=head[v];
    head[v]=tol++;
}
void BFS(int start,int end)
{
    memset(dep,-1,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]=1;
    int que[MAXN];
    int front,rear;
    front=rear=0;
    dep[end]=0;
    que[rear++]=end;
    while(front!=rear)
    {
        int u=que[front++];
        if(front==MAXN)front=0;
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].to;
            if(edge[i].cap!=0||dep[v]!=-1)continue;
            que[rear++]=v;
            if(rear==MAXN)rear=0;
            dep[v]=dep[u]+1;
            ++gap[dep[v]];
        }
    }
}
int SAP(int start,int end)
{
    int res=0;
    BFS(start,end);
    int cur[MAXN];
    int S[MAXN];
    int top=0;
    memcpy(cur,head,sizeof(head));
    int u=start;
    int i;
    while(dep[start]<n)
    {
        if(u==end)
        {
            int temp=INF;
            int inser;
            for(i=0;i<top;i++)
               if(temp>edge[S[i]].cap)
               {
                   temp=edge[S[i]].cap;
                   inser=i;
               }
            for(i=0;i<top;i++)
            {
                edge[S[i]].cap-=temp;
                edge[S[i]^1].cap+=temp;
            }
            res+=temp;
            top=inser;
            u=edge[S[top]].from;
        }
        if(u!=end&&gap[dep[u]-1]==0)//出现断层，无增广路
          break;
        for(i=cur[u];i!=-1;i=edge[i].next)
           if(edge[i].cap!=0&&dep[u]==dep[edge[i].to]+1)
             break;
        if(i!=-1)
        {
            cur[u]=i;
            S[top++]=i;
            u=edge[i].to;
        }
        else
        {
            int min=n;
            for(i=head[u];i!=-1;i=edge[i].next)
            {
                if(edge[i].cap==0)continue;
                if(min>dep[edge[i].to])
                {
                    min=dep[edge[i].to];
                    cur[u]=i;
                }
            }
            --gap[dep[u]];
            dep[u]=min+1;
            ++gap[dep[u]];
            if(u!=start)
              u=edge[S[--top]].from;
        }

    }
    return res;
}

int main()
{
    //freopen("B.in","r",stdin);
    //freopen("B.out","w",stdout);
    int N,M;
    int u,v;
    int start;
    int end;
    while(scanf("%d%d",&N,&M)!=EOF)
    {
        init();
        scanf("%d%d",&start,&end);
        start=2*start-1;
        end=2*end;
        n=2*N;
        for(int i=1;i<=N;i++)
        {
            scanf("%d",&u);
            addedge(2*i-1,2*i,u);
            addedge(2*i,2*i-1,u);
        }
        while(M--)
        {
            scanf("%d%d",&u,&v);
            addedge(2*u,2*v-1,INF);
            addedge(2*v,2*u-1,INF);//这里一定要注意
        }
        printf("%d\n",SAP(start,end));
    }
    return 0;
}