HDU 3879 Base Station

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mob60475702c725 2015-11-18 10:48:00

文章标签 最大权闭合子图网络流最大流最小割 java 文章分类 后端开发

Base Station

Time Limit: 2000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 3879
64-bit integer IO format: %I64d Java class name: Main

A famous mobile communication company is planning to build a new set of base stations. According to the previous investigation, n places are chosen as the possible new locations to build those new stations. However, the condition of each position varies much, so the costs to built a station at different places are different. The cost to build a new station at the ith place is P_i (1<=i<=n).

When complete building, two places which both have stations can communicate with each other.

Besides, according to the marketing department, the company has received m requirements. The ith requirement is represented by three integers A_i, B_i and C_i, which means if place A_iand B_i can communicate with each other, the company will get C_i profit.

Now, the company wants to maximize the profits, so maybe just part of the possible locations will be chosen to build new stations. The boss wants to know the maximum profits.

Input

Multiple test cases (no more than 20), for each test case:
The first line has two integers n (0<n<=5000) and m (0<m<=50000).
The second line has n integers, P1 through Pn, describes the cost of each location.
Next m line, each line contains three integers, A_i, B_i and C_i, describes the ith requirement.

Output

One integer each case, the maximum profit of the company.

Sample Input

Sample Output

Source

2011 Multi-University Training Contest 5 - Host by BNU

解题：最大权闭合子图

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 const int INF = 0x3f3f3f3f;
 4 const int maxn = 100010;
 5 struct arc{
 6     int to,flow,next;
 7     arc(int x = 0,int y = 0,int z = -1){
 8         to = x;
 9         flow = y;
10         next = z;
11     }
12 }e[1000010];
13 int head[maxn],d[maxn],gap[maxn],tot,S,T;
14 void add(int u,int v,int flow){
15     e[tot] = arc(v,flow,head[u]);
16     head[u] = tot++;
17     e[tot] = arc(u,0,head[v]);
18     head[v] = tot++;
19 }
20 queue<int>q;
21 void bfs(){
22     for(int i = 0; i <= T; ++i){
23         d[i] = -1;
24         gap[i] = 0;
25     }
26     d[T] = 0;
27     q.push(T);
28     while(!q.empty()){
29         int u = q.front();
30         q.pop();
31         ++gap[d[u]];
32         for(int i = head[u]; ~i; i = e[i].next){
33             if(d[e[i].to] == -1){
34                 d[e[i].to] = d[u] + 1;
35                 q.push(e[i].to);
36             }
37         }
38     }
39 }
40 int dfs(int u,int low){
41     if(u == T) return low;
42     int tmp = 0,minH = T - 1;
43     for(int i = head[u]; ~i; i = e[i].next){
44         if(e[i].flow && d[e[i].to] + 1 == d[u]){
45             int a = dfs(e[i].to,min(low,e[i].flow));
46             e[i].flow -= a;
47             e[i^1].flow += a;
48             low -= a;
49             tmp += a;
50             if(!low) break;
51             if(d[S] >= T) return tmp;
52         }
53         if(e[i].flow) minH = min(minH,d[e[i].to]);
54     }
55     if(!tmp){
56         if(--gap[d[u]] == 0) d[S] = T;
57         ++gap[d[u] = minH + 1];
58     }
59     return tmp;
60 }
61 int sap(int ret = 0){
62     bfs();
63     while(d[S] < T) ret += dfs(S,INF);
64     return ret;
65 }
66 int main(){
67     int n,m,u,v,w;
68     while(~scanf("%d%d",&n,&m)){
69         memset(head,-1,sizeof head);
70         int sum = tot = 0;
71         S = n + m + 1;
72         T = S + 1;
73         for(int i = 1; i <= n; ++i){
74             scanf("%d",&w);
75             add(S,i,w);
76         }
77         for(int i = 1; i <= m; ++i){
78             scanf("%d%d%d",&u,&v,&w);
79             sum += w;
80             add(u,i + n,INF);
81             add(v,i + n,INF);
82             add(i + n,T,w);
83         }
84         printf("%d\n",sum-sap());
85     }
86     return 0;
87 }