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 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  

 

 

 

拆點+最大流

  1 #include<iostream>
  2 #include<cstring>
  3 #include<string>
  4 #include<cmath>
  5 #include<cstdio>
  6 #include<algorithm>
  7 #include<queue>
  8 #include<vector>
  9 #include<set>
 10 #define maxn 200005
 11 #define MAXN 200005
 12 #define mem(a,b) memset(a,b,sizeof(a))
 13 const int N=200005;
 14 const int M=200005;
 15 const int INF=0x3f3f3f3f;
 16 using namespace std;
 17 int n;
 18 struct Edge{
 19     int v,next;
 20     int cap,flow;
 21 }edge[MAXN*20];//注意這裡要開的夠大。。不然WA在這裡真的想罵人。。問題是還不報RE。。
 22 int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
 23 int cnt=0;//實際儲存總邊數
 24 void isap_init()
 25 {
 26     cnt=0;
 27     memset(pre,-1,sizeof(pre));
 28 }
 29 void isap_add(int u,int v,int w)//加邊
 30 {
 31     edge[cnt].v=v;
 32     edge[cnt].cap=w;
 33     edge[cnt].flow=0;
 34     edge[cnt].next=pre[u];
 35     pre[u]=cnt++;
 36 }
 37 void add(int u,int v,int w){
 38     isap_add(u,v,w);
 39     isap_add(v,u,0);
 40 }
 41 bool bfs(int s,int t)//其實這個bfs可以融合到下面的迭代裡，但是好像是時間要長
 42 {
 43     memset(dep,-1,sizeof(dep));
 44     memset(gap,0,sizeof(gap));
 45     gap[0]=1;
 46     dep[t]=0;
 47     queue<int>q;
 48     while(!q.empty())
 49     q.pop();
 50     q.push(t);//從匯點開始反向建層次圖
 51     while(!q.empty())
 52     {
 53         int u=q.front();
 54         q.pop();
 55         for(int i=pre[u];i!=-1;i=edge[i].next)
 56         {
 57             int v=edge[i].v;
 58             if(dep[v]==-1&&edge[i^1].cap>edge[i^1].flow)//注意是從匯點反向bfs，但應該判斷正向弧的餘量
 59             {
 60                 dep[v]=dep[u]+1;
 61                 gap[dep[v]]++;
 62                 q.push(v);
 63                 //if(v==sp)//感覺這兩句優化加了一般沒錯，但是有的題可能會錯，所以還是註釋出來，到時候視情況而定
 64                 //break;
 65             }
 66         }
 67     }
 68     return dep[s]!=-1;
 69 }
 70 int isap(int s,int t)
 71 {
 72     if(!bfs(s,t))
 73     return 0;
 74     memcpy(cur,pre,sizeof(pre));
 75     //for(int i=1;i<=n;i++)
 76     //cout<<"cur "<<cur[i]<<endl;
 77     int u=s;
 78     path[u]=-1;
 79     int ans=0;
 80     while(dep[s]<n)//迭代尋找增廣路,n為節點數
 81     {
 82         if(u==t)
 83         {
 84             int f=INF;
 85             for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增廣路
 86                 f=min(f,edge[i].cap-edge[i].flow);
 87             for(int i=path[u];i!=-1;i=path[edge[i^1].v])
 88             {
 89                 edge[i].flow+=f;
 90                 edge[i^1].flow-=f;
 91             }
 92             ans+=f;
 93             u=s;
 94             continue;
 95         }
 96         bool flag=false;
 97         int v;
 98         for(int i=cur[u];i!=-1;i=edge[i].next)
 99         {
100             v=edge[i].v;
101             if(dep[v]+1==dep[u]&&edge[i].cap-edge[i].flow)
102             {
103                 cur[u]=path[v]=i;//當前弧優化
104                 flag=true;
105                 break;
106             }
107         }
108         if(flag)
109         {
110             u=v;
111             continue;
112         }
113         int x=n;
114         if(!(--gap[dep[u]]))return ans;//gap優化
115         for(int i=pre[u];i!=-1;i=edge[i].next)
116         {
117             if(edge[i].cap-edge[i].flow&&dep[edge[i].v]<x)
118             {
119                 x=dep[edge[i].v];
120                 cur[u]=i;//常數優化
121             }
122         }
123         dep[u]=x+1;
124         gap[dep[u]]++;
125         if(u!=s)//當前點沒有增廣路則後退一個點
126         u=edge[path[u]^1].v;
127      }
128      return ans;
129 }
130 
131 
132 int main(){
133     std::ios::sync_with_stdio(false);
134     int m,s,t;
135     while(cin>>n>>m){
136         cin>>s>>t;
137         t+=n;
138         int a,b,c;
139         isap_init();
140         for(int i=1;i<=n;i++){
141             cin>>c;
142             add(i,i+n,c);
143         }
144         for(int i=1;i<=m;i++){
145             cin>>a>>b;
146             add(a+n,b,INF);
147             add(b+n,a,INF);
148         }
149         n=n+n;
150         cout<<isap(s,t)<<endl;
151     }
152 }

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