[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;
}

 

posted @ 2017-11-25 15:34  Elpsywk  阅读(191)  评论(0编辑  收藏  举报