POJ 2135 Farm Tour (dinic算法,网络流)
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构图方法:
注意题目中的边为无向边。新建源点s 和 汇点t 每两条道路连一条容量为1,费用为w的边。s到1连一条容量为1,费用为0 的边,n到 t 连一条容量为1,费用为0 的边,求最大流。
#include <iostream>#include <cstring>#include <cstdlib>#include <cstdio>#include <algorithm>#include <queue>#include <vector>#include <cmath>#define LL long longusing namespace std;const int maxn = 10000 + 10;const int INF = 1000000000;struct Edge{ int from,to,cap,flow,cost; Edge(int u,int v,int c,int f,int w):from(u),to(v),cap(c),flow(f),cost(w) { }};int n , m;vector<Edge>edges;vector<int>G[maxn];int inq[maxn];int d[maxn];int p[maxn];int a[maxn];void init(){ for(int i=0;i<=n+1;i++) G[i].clear(); edges.clear();}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)); int M = edges.size(); G[from].push_back(M-2); G[to].push_back(M-1);}bool SPFA(int s,int t,int& flow,LL& cost){ for(int i=0;i<=n+1;i++) d[i] = INF; memset(inq,0,sizeof(inq)); d[s] = 0;inq[s] = 1;p[s] = 0;a[s] = INF; queue<int>Q; Q.push(s); while(!Q.empty()) { int u = Q.front();Q.pop(); inq[u] = 0; 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] = 1;} } } } if(d[t] == INF) return false; flow += a[t]; cost += (LL)d[t] * (LL)a[t]; for(int u=t;u!=s;u=edges[p[u]].from) { edges[p[u]].flow += a[t]; edges[p[u]^1].flow -= a[t]; } return true;}int MincostMaxflow(int s,int t,LL &cost){ int flow = 0;cost = 0; while(SPFA(s,t,flow,cost)); return flow;}int main(){while(scanf("%d%d",&n,&m)!=EOF){init();int u , v , w;for(int i=1;i<=m;i++){scanf("%d%d%d",&u,&v,&w);AddEdge(u,v,1,w);AddEdge(v,u,1,w);}int s = 0 , t = n + 1;AddEdge(s,1,2,0);AddEdge(n,t,2,0);LL cost = 0;int ans = MincostMaxflow(s,t,cost);printf("%lld\n",cost);}return 0;}
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