Cogs 12 运输问题2 （有上下界网络流） 

  1 #include <cstdlib>
  2 #include <algorithm>
  3 #include <cstring>
  4 #include <iostream>
  5 #include <cstdio>
  6 #include <vector>
  7 
  8 using namespace std;
  9 
 10 const int N = 100 +  5;
 11 const int oo = 0x3f3f3f3f;
 12 
 13 struct Edge {
 14   int from, to, cap, flow;
 15 };
 16 
 17 struct Dinic {
 18   int n, m, s, t;
 19   int dis[N], cur[N], que[N << 1];
 20   bool vis[N];
 21   vector <Edge> edges;
 22   vector <int> G[N];
 23 
 24   void add(int from, int to, int cap) {
 25     edges.push_back((Edge) {from, to, cap, 0});
 26     edges.push_back((Edge) {to, from, 0, 0});
 27     m = edges.size();
 28     G[from].push_back(m - 2);
 29     G[to].push_back(m - 1);
 30   }
 31 
 32   bool bfs() {
 33     int head = 1, tail = 1;
 34 
 35     memset(vis, false, sizeof vis);
 36     dis[s] = 0; vis[s] = true; que[head] = s;
 37     while(head <= tail) {
 38       int x = que[head];
 39 
 40       for(int i = 0; i < (signed) G[x].size(); ++ i) {
 41         Edge &e = edges[G[x][i]];
 42 
 43         if(!vis[e.to] && e.cap > e.flow) {
 44           vis[e.to] = true;
 45           dis[e.to] = dis[x] + 1;
 46           que[++ tail] = e.to;
 47         }
 48       }
 49       ++ head;
 50     }
 51     return vis[t];
 52   }
 53 
 54   int dfs(int x, int a) {
 55     if(x == t || a == 0) return a;
 56 
 57     int flw = 0, f;
 58 
 59     for(int &i = cur[x]; i < (signed) G[x].size(); ++ i) {
 60       Edge &e = edges[G[x][i]];
 61       
 62       if(dis[e.to] == dis[x] + 1 && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0) {
 63         e.flow += f; edges[G[x][i] ^ 1].flow -= f; flw += f; a -= f;
 64         if(a == 0) break;
 65       }
 66     }
 67     return flw;
 68   }
 69 
 70   int MaxFlow(int s, int t) {
 71     this->s = s; this->t = t;
 72 
 73     int flw = 0;
 74 
 75     while(bfs()) {
 76       memset(cur, 0, sizeof cur);
 77       flw += dfs(s, oo);
 78     }
 79     return flw;
 80   }
 81 }net;
 82 
 83 int n, M[N];
 84 
 85 int main() {
 86 #ifndef ONLINE_JUDGE
 87   freopen("maxflowb.in", "r", stdin);
 88   freopen("maxflowb.out", "w", stdout);
 89 #endif
 90 
 91   int Up, Down;
 92   
 93   scanf("%d", &n); net.n = n + 1;
 94   for(int i = 1; i <= n; ++ i) {
 95     for(int j = 1; j <= n; ++ j) {
 96       scanf("%d%d", &Down, &Up);
 97       M[i] -= Down; M[j] += Down;
 98       net.add(i, j, Up - Down);
 99     }
100   }
101   net.add(n, 1, oo);
102   for(int i = 1; i <= n; ++ i) {
103     if(M[i] > 0) {
104       net.add(0, i, M[i]);
105     }
106     else if(M[i] < 0){
107       net.add(i, n + 1, -M[i]);
108     }
109   }
110   net.MaxFlow(0, n + 1);
111   printf("%d\n", net.MaxFlow(1, n));
112   
113 #ifndef ONLINE_JUDGE
114   fclose(stdin); fclose(stdout);
115 #endif
116   return 0;
117 }
Cogs 12

连边方式见图。

posted @ 2016-03-16 20:20  漫步者。！~  阅读(189)  评论(0)  编辑  收藏  举报