poj 2135 Farm Tour

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http://poj.org/problem?id=2135

此题中的费用是路径长度，流量是经过的点的数目（流量越大对应于更好地参观农场），因此是一个最小费用最大流问题，使用SPFA来求解。

AC代码：

#include <queue>#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;const int MAXN = 1010;const int MAXM = 1000010;const int INF = 0x3f3f3f3f;int sumFlow,S,T,n,m;int NE,head[MAXN],dis[MAXN],pre[MAXN],vis[MAXN];struct edge{int u,v,c,w;int next;}node[MAXM<<2];void Init(){NE = 0;memset(head, -1, sizeof(head));}void addedge(int u, int v, int c, int w){//前向弧与后向弧node[NE].v = v;node[NE].u = u;node[NE].c = c;node[NE].w = w;node[NE].next = head[u];head[u] = NE++;        node[NE].v = u;node[NE].u = v;node[NE].c = 0;node[NE].w = -w;node[NE].next = head[v];head[v] = NE++;}bool SPFA(){queue <int> Q;memset(vis, 0, sizeof(vis));memset(pre, -1, sizeof(pre));memset(dis, 0x3f, sizeof(dis));vis[S] = 1, dis[S] = 0;Q.push(S);while(!Q.empty()){int u = Q.front();Q.pop();vis[u] = 0;for(int i=head[u]; i!=-1; i=node[i].next){int v = node[i].v;if(node[i].c > 0 && dis[v] > dis[u] + node[i].w){dis[v] = dis[u] + node[i].w;pre[v] = i;if(!vis[v]){Q.push(v);vis[v] = 1;}}}}if(dis[T] == INF) return false;return true;}int MCMF(){int flow = 0;int mincost = 0;while(SPFA()){int minflow = INF;for(int i=pre[T]; i!=-1; i=pre[node[i].u])if(node[i].c < minflow)minflow = node[i].c;flow += minflow;for(int i=pre[T]; i!=-1; i=pre[node[i].u]){node[i].c -= minflow;node[i^1].c += minflow;}mincost += dis[T] * minflow;}sumFlow = flow;return mincost;}int main(){while(scanf("%d%d",&n,&m) == 2){Init();S = 0, T = n+1;while(m--){int u,v,w;scanf("%d%d%d",&u,&v,&w);addedge(u,v,1,w);addedge(v,u,1,w);}addedge(S,1,2,0);addedge(n,T,2,0);int ans = MCMF();printf("%d\n",ans);}return 0;}

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