hdu4289（拆点最大流）

Control

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2564    Accepted Submission(s): 1095

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 10⁷.
　　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

题意：有N个城市，现在城市S出现了一伙歹徒，他们想运送一些炸弹到D城市，不过警方已经得到了线报知道他们的事情，不过警察不知道他们所在的具体位置，所以只能采取封锁城市的办法来阻断暴徒，不过封锁城市是需要花费一定代价的，由于警局资金比较紧张，所以想知道如果完全阻断暴徒从S城市到达D城市的最小需要花费的代价。

思路：将每个点都拆分成2个，表示为i和i*,i是原来的起点，然后i到i*的权值为i点原来的权值，连边的时候要注意的是要从i*连接i，因为最初在起点的时候，首先要走的就是从i到i*，然后接下来一定是从i*出发回到i这一边，假设是j，然后j走j*，这样循环往复，所以边的方向都应该是i*到i。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
using namespace std;
typedef long long ll;
const   int oo=1e9;
/**oo 表示无穷大*/
const  int mm=111111111;
/**mm 表示边的最大数量，记住要是原图的两倍，在加边的时候都是双向的*/
const  int mn=2010;
/**mn 表示点的最大数量*/
int node,src,dest,edge;
/**node 表示节点数，src 表示源点，dest 表示汇点，edge 统计边数*/
int ver[mm],flow[mm],nex[mm];
/**ver 边指向的节点，flow 边的容量 ，next 链表的下一条边*/
int head[mn],work[mn],dis[mn],q[mn];
void prepare(int _node, int _src,int _dest)
{
    node=_node,src=_src,dest=_dest;
    for(int i=0; i<node; ++i)head[i]=-1;
    edge=0;
}
/**增加一条 u 到 v 容量为 c 的边*/
void addedge( int u,  int v,  int c)
{
    ver[edge]=v,flow[edge]=c,nex[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,nex[edge]=head[v],head[v]=edge++;
}
/**广搜计算出每个点与源点的最短距离，如果不能到达汇点说明算法结束*/
bool Dinic_bfs()
{
    int i,u,v,l,r=0;
    for(i=0; i<node; ++i)dis[i]=-1;
    dis[q[r++]=src]=0;
    for(l=0; l<r; ++l)
        for(i=head[u=q[l]]; i>=0; i=nex[i])
            if(flow[i]&&dis[v=ver[i]]<0)
            {
                /**这条边必须有剩余容量*/
                dis[q[r++]=v]=dis[u]+1;
                if(v==dest)  return 1;
            }
    return 0;
}
/**寻找可行流的增广路算法，按节点的距离来找，加快速度*/
int Dinic_dfs(  int u, int exp)
{
    if(u==dest)  return exp;
    /**work 是临时链表头，这里用 i 引用它，这样寻找过的边不再寻找*/
    for(  int &i=work[u],v,tmp; i>=0; i=nex[i])
        if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
        {
            flow[i]-=tmp;
            flow[i^1]+=tmp;
            /**正反向边容量改变*/
            return tmp;
        }
    return 0;
}

int Dinic_flow()
{
    int i,ret=0,delta;
    while(Dinic_bfs())
    {
        for(i=0; i<node; ++i)work[i]=head[i];
        while((delta=Dinic_dfs(src,oo)))ret+=delta;
    }
    return ret;
}

int main()
{
    int n,m,s,t;
    while(~scanf("%d%d",&n,&m))
    {
        scanf("%d%d",&s,&t);
        prepare(2*n+1,s,t+n);
        for(int i=1; i<=n; i++)
        {
            int x;
            scanf("%d",&x);
            addedge(i,i+n,x);
        }
        for(int i=0; i<m; i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            addedge(a+n,b,oo);
            addedge(b+n,a,oo);
        }
        printf("%d\n",Dinic_flow());
    }
}