POJ 2135 Farm Tour(最小费用流)

//最小费用流 //构图关键是添加超级源点，超级汇点 //题意是要找一条可以再终点往返的路，这2条路距离之和最短且每条路只能经过一次 //第一步将问题转化为找2条从起点到终点 //第二步与费用流挂钩，将距离当做费用，将容量设置为1是为了保证每条路只经过一次 //超级源点和起点的容量为2，终点和超级汇点容量为2，使得最多找2条路 //最终的mincost即为答案 //注意双向边这个条件，当出现双向边或多重边时，就得用邻接表储存了 #include<iostream> #include<queue> using namespace std; const int MAXN = 1003; const int MAXM = 40005; const int INF = 1000000000; int mincost,maxflow; int N,M; int U[MAXM],V[MAXM],cap[MAXM],flow[MAXM],cost[MAXM],next[MAXM]; int head[MAXN],pre[MAXN],Edge[MAXN],dis[MAXN]; int m; void addEdge(int u,int v,int Cap,int Cost) { flow[m] = flow[m+1] = 0; V[m] = v; U[m] = u; cap[m] = Cap; cost[m] = Cost; next[m] = head[u]; head[u] = m++; V[m] = u; U[m] = v; cap[m] = 0; cost[m] = -Cost; next[m] = head[v]; head[v] = m++; } void buildGraph() { scanf("%d%d",&N,&M); int u,v,c; m = 0; memset(head,-1,sizeof(head)); while(M--) { scanf("%d%d%d",&u,&v,&c); addEdge(u,v,1,c); addEdge(v,u,1,c); } addEdge(0,1,2,0); addEdge(N,N+1,2,0); } void MCMF(int st,int ed) { queue<int> q; memset(flow,0,sizeof(flow)); mincost = maxflow = 0; for(;;) { bool inq[MAXN]; for(int i = st;i <= ed;++i) dis[i] = (i == st ? 0 : INF); memset(inq,0,sizeof(inq)); q.push(st); while(!q.empty()) { int u = q.front(); q.pop(); inq[u] = 0; for(int e = head[u];e != -1;e = next[e]) { if(cap[e] > flow[e] && dis[u] + cost[e] < dis[V[e]]) { dis[V[e]] = dis[u] + cost[e]; pre[V[e]] = U[e]; Edge[V[e]] = e; if(!inq[V[e]]) { q.push(V[e]); inq[V[e]] = 1; } } } }//SPFA增广 if(dis[ed] == INF) break; int delta = INF;//delta为可改进量 for(int u = ed;u != st;u = pre[u]) delta = min(delta,cap[Edge[u]] - flow[Edge[u]]);//遍历最短路径的边，并修改可改进量 for(int u = ed;u != st;u = pre[u]) { flow[Edge[u]] += delta;//更新正向流量 flow[Edge[u] ^ 1] -= delta;//通过异或1取得反向边的序号，并更新反向流量 } mincost += dis[ed] * delta; maxflow += delta; } } int main() { //freopen("in.txt","r",stdin); buildGraph(); MCMF(0,N+1); printf("%d/n",mincost); return 0; }