HDU 3435 A new Graph Game(费用流)
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题意:给你一个N个节点M条边的无向图,要你求该图有1个或多个不相交有向环构成时,所有这些有向环的最小权值.
思路:和HDU1853差不多,只是变成了无向图,那么把一条无向边变成两条相反方向的有向边即可
注意:有重边,要处理一下
#include<cstdio>#include<cstring>#include<queue>#include<algorithm>#include<vector>#define INF 1e9using namespace std;const int maxn=2500;int T,cas=1;struct Edge{ int from,to,cap,flow,cost; Edge(){} Edge(int f,int t,int c,int fl,int co):from(f),to(t),cap(c),flow(fl),cost(co){}};struct MCMF{ int n,m,s,t; vector<Edge> edges; vector<int> G[maxn]; bool inq[maxn]; //是否在队列 int d[maxn]; //Bellman_ford单源最短路径 int p[maxn]; //p[i]表从s到i的最小费用路径上的最后一条弧编号 int a[maxn]; //a[i]表示从s到i的最小残量 //初始化 void init(int n,int s,int t) { this->n=n, this->s=s, this->t=t; edges.clear(); for(int i=0;i<n;++i) G[i].clear(); } //添加一条有向边 void AddEdge(int from,int to,int cap,int cost) { edges.push_back(Edge(from,to,cap,0,cost)); edges.push_back(Edge(to,from,0,0,-cost)); m=edges.size(); G[from].push_back(m-2); G[to].push_back(m-1); } //求一次增广路 bool BellmanFord(int &flow, int &cost) { for(int i=0;i<n;++i) d[i]=INF; memset(inq,0,sizeof(inq)); d[s]=0, a[s]=INF, inq[s]=true, p[s]=0; queue<int> Q; Q.push(s); while(!Q.empty()) { int u=Q.front(); Q.pop(); inq[u]=false; for(int i=0;i<G[u].size();++i) { Edge &e=edges[G[u][i]]; if(e.cap>e.flow && d[e.to]>d[u]+e.cost) { d[e.to]= d[u]+e.cost; p[e.to]=G[u][i]; a[e.to]= min(a[u],e.cap-e.flow); if(!inq[e.to]){ Q.push(e.to); inq[e.to]=true; } } } } if(d[t]==INF) return false; flow +=a[t]; cost +=a[t]*d[t]; int u=t; while(u!=s) { edges[p[u]].flow += a[t]; edges[p[u]^1].flow -=a[t]; u = edges[p[u]].from; } return true; } //求出最小费用最大流 int Min_cost(int num) { int flow=0,cost=0; while(BellmanFord(flow,cost)); return num==flow?cost:-1; }}mc;int d[maxn][maxn];int main(){int n,m;scanf("%d",&T); while (T--){ scanf("%d%d",&n,&m); mc.init(n*2+2,0,n*2+1);memset(d,0,sizeof(d));for (int i=1;i<=n;i++){mc.AddEdge(0,i,1,0);mc.AddEdge(n+i,n*2+1,1,0);}for (int i = 1;i<=m;i++){int u,v,w;scanf("%d%d%d",&u,&v,&w);if (d[u][v] > w || !d[u][v]){d[u][v]=w;d[v][u]=w;}}for (int i = 1;i<=n;i++)for (int j = i+1;j<=n;j++)if (d[i][j]!=0){mc.AddEdge(i,j+n,1,d[i][j]); mc.AddEdge(j,i+n,1,d[i][j]);} int ans = mc.Min_cost(n);if (ans == -1)printf("Case %d: NO\n",cas++);else printf("Case %d: %d\n",cas++,ans);}}
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