[poj2135]Farm Tour(最小费用流)
解题关键:最小费用流
代码一:bellma-ford $O(FVE)$ bellman-ford求最短路,并在最短路上增广,速度较慢
#include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cmath> #include<vector> #define inf 0x3f3f3f3f #define MAX_V 10010 using namespace std; typedef long long ll; struct edge{int to,cap,cost,rev;}; int V; vector<edge>G[MAX_V]; int dist[MAX_V]; int prevv[MAX_V],preve[MAX_V]; void add_edge(int from,int to,int cap,int cost){ G[from].push_back((edge){to,cap,cost,G[to].size()}); G[to].push_back((edge){from,0,-cost,G[from].size()-1}); } int min_cost_flow(int s,int t,int f){ int res=0; while(f>0){ fill(dist,dist+V,inf); dist[s]=0; bool update=true; while(update){ update=false; for(int v=0;v<V;v++){ if(dist[v]==inf) continue; for(int i=0;i<G[v].size();i++){ edge &e=G[v][i]; if(e.cap>0&&dist[e.to]>dist[v]+e.cost){ dist[e.to]=dist[v]+e.cost; prevv[e.to]=v; preve[e.to]=i; update=true; } } } } if(dist[t]==inf) return -1; int d=f; for(int v=t;v!=s;v=prevv[v]) d=min(d,G[prevv[v]][preve[v]].cap); f-=d; res+=d*dist[t]; for(int v=t;v!=s;v=prevv[v]){ edge &e=G[prevv[v]][preve[v]]; e.cap-=d; G[v][e.rev].cap+=d; } } return res; } int n,m,t1,t2,t3; int main(){ while(scanf("%d%d",&n,&m)!=EOF){ memset(G,0,sizeof G); V=n; for(int i=0;i<m;i++){ scanf("%d%d%d",&t1,&t2,&t3); add_edge(t1-1,t2-1,1,t3); add_edge(t2-1,t1-1,1,t3); } printf("%d\n",min_cost_flow(0,n-1,2)); } return 0; }
代码二:dijkstra,$O(FElogV)$
这里是通过一个定理
s到v的最短距离<=s到u的最短距离+dis(e)
s到u的最短距离+dis(e)-s到v的最短距离>=0
将原先的距离转化为上述的等效距离,即可保证图中无负权边,所以可以用dijkstra算法堆优化来保证复杂度。
#include<cstdio> #include<cstring> #include<algorithm> #include<cstdlib> #include<iostream> #include<cmath> #include<vector> #include<queue> #define inf 0x3f3f3f3f #define MAX_V 10010 using namespace std; typedef long long ll; typedef pair<int,int>P; struct edge{int to,cap,cost,rev;}; int V; vector<edge>G[MAX_V]; int h[MAX_V],dist[MAX_V],prevv[MAX_V],preve[MAX_V]; void add_edge(int from,int to,int cap,int cost){ G[from].push_back((edge){to,cap,cost,G[to].size()}); G[to].push_back((edge){from,0,-cost,G[from].size()-1}); } int min_cost_flow(int s,int t,int f){ int res=0; fill(h,h+V,0); while(f>0){ priority_queue<P,vector<P>,greater<P> >que; fill(dist,dist+V,inf); dist[s]=0; que.push(P(0,s)); while(!que.empty()){ P p=que.top();que.pop(); int v=p.second; if(dist[v]<p.first) continue; for(int i=0;i<G[v].size();i++){ edge &e=G[v][i]; if(e.cap>0&&dist[e.to]>dist[v]+e.cost+h[v]-h[e.to]){ dist[e.to]=dist[v]+e.cost+h[v]-h[e.to]; prevv[e.to]=v; preve[e.to]=i; que.push(P(dist[e.to],e.to)); } } } if(dist[t]==inf) return -1; for(int v=0;v<V;v++) h[v]+=dist[v]; //增广 int d=f; for(int v=t;v!=s;v=prevv[v]) d=min(d,G[prevv[v]][preve[v]].cap); f-=d; res+=d*h[t]; for(int v=t;v!=s;v=prevv[v]){ edge &e=G[prevv[v]][preve[v]]; e.cap-=d; G[v][e.rev].cap+=d; } } return res; } int n,m,t1,t2,t3; int main(){ while(scanf("%d%d",&n,&m)!=EOF){ memset(G,0,sizeof G); V=n; for(int i=0;i<m;i++){ scanf("%d%d%d",&t1,&t2,&t3); add_edge(t1-1,t2-1,1,t3); add_edge(t2-1,t1-1,1,t3); } printf("%d\n",min_cost_flow(0,n-1,2)); } return 0; }