HDU4411 最小费用流

题目：http://acm.hdu.edu.cn/showproblem.php?pid=4411

  1. floyd处理出最短路
  2. 每个点拆为i、i+n，i到i+n连一条容量为1，费用为负无穷的边，代表这个城市必须访问
  3. i+n到j（j>i)分别建边，容量为1，费用为i、j最短路
  4. i+n到汇点建立容量为1费用为g[0][i]的边，代表会警察局
  5. 0到i建立容量为1费用为g[0][i]的边
  6. 0到汇点建立容量为k，费用为0的边，代表有的警察可能一直待在警察局
  7. 源点到0建立容量为k费用为0的边

ps：这题会有重边，以后一定要注意，每道题都要判重！

 1 #include<bits/stdc++.h> 
 2 using namespace std;
 3 const int inf = 0x3f3f3f3f;
 4 const int maxv = 222;
 5 typedef pair<int,int> P;
 6 struct edge{
 7     int to,cap,cost,rev;
 8 };
 9 int V;
10 vector<edge> g[maxv];
11 int h[maxv];
12 int dist[maxv];
13 int prevv[maxv],preve[maxv];
14 void addedge(int from,int to,int cap,int cost){
15     g[from].push_back(edge{to,cap,cost,g[to].size()});
16     g[to].push_back((edge{from,0,-cost,g[from].size()-1}));
17 } 
18 int solve(int s,int t,int f){
19     int res = 0;
20     memset(h,0,sizeof(h)) ;
21     while(f > 0){
22         priority_queue<P,vector<P>,greater<P> > que;
23         memset(dist,inf,sizeof(dist));
24         dist[s] = 0;
25         que.push(P(0,s));
26         while(!que.empty()){
27             P p = que.top();
28             que.pop();
29             int v = p.second;
30             if(dist[v] < p.first)
31                 continue;
32             for(int i = 0;i<g[v].size();i++){
33                 edge &e = g[v][i] ;
34                 if(e.cap>0 && dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){
35                     dist[e.to] = dist[v]+e.cost+h[v]-h[e.to];
36                     prevv[e.to] = v;
37                     preve[e.to] = i;
38                     que.push(P(dist[e.to],e.to));
39                 }
40             }
41         }
42         if(dist[t] == inf)
43             return -1;
44         for(int i = 0;i<V;i++)
45             h[i] += dist[i];
46         int d = f;
47         for(int v = t;v!=s;v=prevv[v])
48             d = min(d,g[prevv[v]][preve[v]].cap);
49         f -= d;
50         res += d*h[t];
51         for(int v = t;v!=s;v = prevv[v]){
52             edge &e = g[prevv[v]][preve[v]];
53             e.cap -= d;
54             g[v][e.rev].cap += d;
55         }
56     }
57     return res;
58 }
59 int mp[111][111];
60 int main(){
61     int n,m,k;
62     while(scanf("%d%d%d",&n,&m,&k)){
63         if(!n && !m && !k)
64             break;
65         V = 2*n+3;
66         int s = 2*n+1,t = 2*n+2;
67         for(int i = 0;i<2*n+3;i++)
68             g[i].clear();
69         memset(mp,inf,sizeof(mp));
70         for(int i = 0;i<=n;i++)
71             mp[i][i] = 0;
72         int u,v,w;
73         while(m--){
74             scanf("%d%d%d",&u,&v,&w);
75             mp[u][v] = mp[v][u] = min(mp[u][v],w);
76         }
77         for(int k = 0;k<=n;k++)
78             for(int i = 0;i<=n;i++)
79                 for(int j = 0;j<=n;j++)
80                     mp[i][j] = min(mp[i][j],mp[i][k]+mp[k][j]);
81         for(int i = 1;i<=n;i++){
82             addedge(0,i,1,mp[0][i]);
83             addedge(i+n,t,1,mp[0][i]);
84             addedge(i,i+n,1,-111111);
85             for(int j = i+1;j<=n;j++){
86                 if(mp[i][j] != inf)
87                     addedge(i+n,j,k,mp[i][j]);
88             }
89         }
90         addedge(s,0,k,0);
91         addedge(0,t,k,0);
92         cout << solve(s,t,k)+n*111111 << endl;
93     }
94     return 0;
95 }

 

posted @ 2016-11-03 17:06  张秦遥  阅读(211)  评论(0)  编辑  收藏  举报