Control（拆点+最大流）

Control

http://acm.hdu.edu.cn/showproblem.php?pid=4289

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4995    Accepted Submission(s): 2057

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

 

Source

2012 ACM/ICPC Asia Regional Chengdu Online

 

拆点+最大流

 #include<iostream>
 #include<cstring>
 #include<string>
 #include<cmath>
 #include<cstdio>
 #include<algorithm>
 #include<queue>
 #include<vector>
 #include<set>
 #define maxn 200005
 #define MAXN 200005
 #define mem(a,b) memset(a,b,sizeof(a))
 const int N=;
 const int M=;
 const int INF=0x3f3f3f3f;
 using namespace std;
 int n;
 struct Edge{
     int v,next;
     int cap,flow;
 }edge[MAXN*];//注意这里要开的够大。。不然WA在这里真的想骂人。。问题是还不报RE。。
 int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
 int cnt=;//实际存储总边数
 void isap_init()
 {
     cnt=;
     memset(pre,-,sizeof(pre));
 }
 void isap_add(int u,int v,int w)//加边
 {
     edge[cnt].v=v;
     edge[cnt].cap=w;
     edge[cnt].flow=;
     edge[cnt].next=pre[u];
     pre[u]=cnt++;
 }
 void add(int u,int v,int w){
     isap_add(u,v,w);
     isap_add(v,u,);
 }
 bool bfs(int s,int t)//其实这个bfs可以融合到下面的迭代里，但是好像是时间要长
 {
     memset(dep,-,sizeof(dep));
     memset(gap,,sizeof(gap));
     gap[]=;
     dep[t]=;
     queue<int>q;
     while(!q.empty())
     q.pop();
     q.push(t);//从汇点开始反向建层次图
     while(!q.empty())
     {
         int u=q.front();
         q.pop();
         for(int i=pre[u];i!=-;i=edge[i].next)
         {
             int v=edge[i].v;
             if(dep[v]==-&&edge[i^].cap>edge[i^].flow)//注意是从汇点反向bfs，但应该判断正向弧的余量
             {
                 dep[v]=dep[u]+;
                 gap[dep[v]]++;
                 q.push(v);
                 //if(v==sp)//感觉这两句优化加了一般没错，但是有的题可能会错，所以还是注释出来，到时候视情况而定
                 //break;
             }
         }
     }
     return dep[s]!=-;
 }
 int isap(int s,int t)
 {
     if(!bfs(s,t))
     return ;
     memcpy(cur,pre,sizeof(pre));
     //for(int i=1;i<=n;i++)
     //cout<<"cur "<<cur[i]<<endl;
     int u=s;
     path[u]=-;
     int ans=;
     while(dep[s]<n)//迭代寻找增广路,n为节点数
     {
         if(u==t)
         {
             int f=INF;
             for(int i=path[u];i!=-;i=path[edge[i^].v])//修改找到的增广路
                 f=min(f,edge[i].cap-edge[i].flow);
             for(int i=path[u];i!=-;i=path[edge[i^].v])
             {
                 edge[i].flow+=f;
                 edge[i^].flow-=f;
             }
             ans+=f;
             u=s;
             continue;
         }
         bool flag=false;
         int v;
         for(int i=cur[u];i!=-;i=edge[i].next)
         {
             v=edge[i].v;
             if(dep[v]+==dep[u]&&edge[i].cap-edge[i].flow)
             {
                 cur[u]=path[v]=i;//当前弧优化
                 flag=true;
                 break;
             }
         }
         if(flag)
         {
             u=v;
             continue;
         }
         int x=n;
         if(!(--gap[dep[u]]))return ans;//gap优化
         for(int i=pre[u];i!=-;i=edge[i].next)
         {
             if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x)
             {
                 x=dep[edge[i].v];
                 cur[u]=i;//常数优化
             }
         }
         dep[u]=x+;
         gap[dep[u]]++;
         if(u!=s)//当前点没有增广路则后退一个点
         u=edge[path[u]^].v;
      }
      return ans;
 }

 int main(){
     std::ios::sync_with_stdio(false);
     int m,s,t;
     while(cin>>n>>m){
         cin>>s>>t;
         t+=n;
         int a,b,c;
         isap_init();
         for(int i=;i<=n;i++){
             cin>>c;
             add(i,i+n,c);
         }
         for(int i=;i<=m;i++){
             cin>>a>>b;
             add(a+n,b,INF);
             add(b+n,a,INF);
         }
         n=n+n;
         cout<<isap(s,t)<<endl;
     }
 }