POJ 2315 最小费用最大流
从1走到N然后从N走回来的最短路程是多少? 转换为费用流来建模.
1: /**
2: 因为e ==0 所以 pe[v] pe[v]^1 是两条相对应的边
3: E[pe[v]].c -= aug; E[pe[v]^1].c += aug;
4:
5: */
6: #include <queue>
7: #include <iostream>
8: #include <string.h>
9: #include <stdio.h>
10: #include <map>
11: using namespace std;
12: #define V 30010 // vertex
13: #define E 150010 // edge
14: #define INF 0x3F3F3F3F
15: struct MinCostMaxFlow
16: {
17: struct Edge
18: {
19: int v, next, cap, cost; // cap 为容量, cost为单位流量费用
20: } edge[E];
21:
22: int head[V], pe[V], pv[V]; // 每个节点的第一条edge[idx]的编号.
23: int dis[V]; // the shortest path to the src
24: bool vis[V]; // visted
25: // pe[v] 存放在增广路上到达v的边(u,v) 在edge[]的位置
26: // pv[u] 存放在增广路上从u出发的边(u,v) 在edge[]中的位置
27: int e, src, sink; // the index of the edge
28: void addedge(int u, int v, int cap, int cost)
29: {
30: edge[e].v = v, edge[e].cap = cap;
31: edge[e].cost = cost,edge[e].next = head[u], head[u] = e++;
32: edge[e].v = u, edge[e].cap = 0;
33: edge[e].cost = -1*cost, edge[e].next = head[v], head[v] = e++;
34: }
35: // 求最短路,不存在负环的时候,比 DFS快
36: int SPFABFS()
37: {
38: memset(vis, 0 ,sizeof(vis));
39: memset(pv, -1, sizeof(pv));
40: for(int i=0; i<V; i++) dis[i] = INF;
41: queue<int> Q;
42: Q.push(src), vis[src] = 1, dis[src] = 0;
43: while(!Q.empty())
44: {
45: int u = Q.front();
46: Q.pop(), vis[u] = 0;
47: for(int i=head[u]; i!=-1; i=edge[i].next)
48: {
49: int v = edge[i].v;
50: if(edge[i].cap > 0 && dis[v] > dis[u] + edge[i].cost )
51: {
52: dis[v] = dis[u] + edge[i].cost;
53: if(!vis[v]) Q.push(v),vis[v] = 1;
54: pv[v] = u, pe[v] = i;
55: }
56: }
57: }
58: if(dis[sink] == INF) return -2; // can't from src to sink.
59: return dis[sink];
60: }
61:
62: pair<int,int> MCMF()
63: {
64: int maxflow = 0, mincost = 0;
65: while(SPFABFS())
66: {
67: if(pv[sink] == -1) break;
68: int aug = INF;
69: for(int i= sink; i!= src; i = pv[i])
70: aug =min(aug, edge[pe[i]].cap);
71: maxflow += aug;
72: mincost += aug * dis[sink];
73: for(int i = sink; i!= src; i = pv[i])
74: {
75: edge[pe[i]].cap -= aug;
76: edge[pe[i]^1].cap += aug;
77: }
78: }
79: return make_pair(maxflow, mincost);
80: }
81:
82: /// 一定需要初始化的是 src, sink
83: void solve()
84: {
85: int N,M;
86: while(scanf("%d%d", &N , &M)!= EOF)
87: {
88: e=0;
89: memset(head, -1,sizeof(head));
90: for(int i=0; i<M; i++)
91: {
92: int a, b, c;
93: scanf("%d%d%d", &a, &b, &c);
94: addedge(a, b,1,c);
95: addedge(b, a,1, c);
96: }
97: src = 0;
98: sink = N+1;
99: addedge(src,1, 2, 0);
100: addedge(N, sink, 2, 0);
101: pair<int,int> ret = MCMF();
102: cout<<ret.second<<endl;
103: }
104: }
105: } mcmf;
106:
107: int main()
108: {
109: mcmf.solve();
110: return 0;
111: }