# 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

拆点+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。。
22 int cur[MAXN],pre[MAXN],gap[MAXN],path[MAXN],dep[MAXN];
23 int cnt=0;//实际存储总边数
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)//加边
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){
40 }
41 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
42 {
43     memset(dep,-1,sizeof(dep));
44     memset(gap,0,sizeof(gap));
45     gap=1;
46     dep[t]=0;
47     queue<int>q;
48     while(!q.empty())
49     q.pop();
50     q.push(t);//从汇点开始反向建层次图
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)//注意是从汇点反向bfs，但应该判断正向弧的余量
59             {
60                 dep[v]=dep[u]+1;
61                 gap[dep[v]]++;
62                 q.push(v);
63                 //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
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)//迭代寻找增广路,n为节点数
81     {
82         if(u==t)
83         {
84             int f=INF;
85             for(int i=path[u];i!=-1;i=path[edge[i^1].v])//修改找到的增广路
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;//当前弧优化
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优化
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;//常数优化
121             }
122         }
123         dep[u]=x+1;
124         gap[dep[u]]++;
125         if(u!=s)//当前点没有增广路则后退一个点
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;
143         }
144         for(int i=1;i<=m;i++){
145             cin>>a>>b;
148         }
149         n=n+n;
150         cout<<isap(s,t)<<endl;
151     }
152 }```
View Code
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• 原文地址：https://www.cnblogs.com/Fighting-sh/p/9949204.html