POJ2135 Farm Tour(最小費用最大流裸題)
阿新 • • 發佈:2019-01-06
Source Code #include<queue> #include<cstdio> #include<string> #include<cstring> #include<iostream> #include<algorithm> using namespace std; const int MAXN = 100000; const int MAXM = 100000; const int INF = 1999999999; struct Edge { int to,next,cap,flow,cost; }edge[MAXM]; int head[MAXN],tol; int pre[MAXN],dis[MAXN]; bool vis[MAXN]; int N;//節點總個數,節點編號從0~N-1 void init(int n) { N = n; tol = 0; memset(head,-1,sizeof(head)); } void addedge(int u,int v,int cap,int cost) { edge[tol].to = v; edge[tol].cap = cap; edge[tol].cost = cost; edge[tol].flow = 0; edge[tol].next = head[u]; head[u] = tol++; edge[tol].to = u; edge[tol].cap = 0; edge[tol].cost = -cost; edge[tol].flow = 0; edge[tol].next = head[v]; head[v] = tol++; } bool spfa(int s,int t) { queue<int>q; for(int i = 0;i <= N;i++) { dis[i] = INF; vis[i] = false; pre[i] = -1; } dis[s] = 0; vis[s] = true; q.push(s); while(!q.empty()) { //cout<<1<<endl; int u = q.front(); q.pop(); vis[u] = false; for(int i = head[u]; i != -1;i = edge[i].next) { int v = edge[i].to; if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost ) { dis[v] = dis[u] + edge[i].cost; pre[v] = i; if(!vis[v]) { vis[v] = true; q.push(v); } } } } if(pre[t] == -1)return false; else return true; } //返回的是最大流,cost存的是最小費用 int minCostMaxflow(int s,int t,int &cost) { int flow = 0; cost = 0; while(spfa(s,t)) { //cout<<1<<endl; int Min = INF; for(int i = pre[t];i != -1;i = pre[edge[i^1].to]) { if(Min > edge[i].cap - edge[i].flow) Min = edge[i].cap - edge[i].flow; } for(int i = pre[t];i != -1;i = pre[edge[i^1].to]) { edge[i].flow += Min; edge[i^1].flow -= Min; cost += edge[i].cost * Min; } flow += Min; } return flow; } int main() { int n,m; while(cin>>n>>m) { init(n+1); int u,v,w,s,t; while(m--) { cin>>u>>v>>w; addedge(u,v,1,w); addedge(v,u,1,w); } addedge(0,1,2,0); addedge(n,n+1,2,0); int cost; int ans=minCostMaxflow(0,n+1,cost); cout<<cost<<endl; } }