最小费用流模板(Bellman-Ford算法找最短路)

来源：互联网发布：数据录入查询系统源码编辑：程序博客网时间：2024/05/16 15:34
模板来自《挑战程序设计竞赛》
poj 3068
#include <stdio.h>#include <string.h>#include <vector>#include <algorithm>using namespace std;struct edge{    int to,cap,cost,rev;    edge(int _to = 0, int _cap = 0, int _cost = 0, int _rev = 0)    :to(_to),cap(_cap),cost(_cost),rev(_rev){}};const int INF = 0x3f3f3f3f;const int MAX_V = 2222;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)    {        memset(dist,0x3f,sizeof(dist));        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 main(){    int n,m,u,v,cost,cnt = 0;    while(scanf("%d %d",&n,&m) && n+m)    {        V = n;        ++cnt;        for(int i = 0; i <= n; ++i) G[i].clear();        for(int i = 0; i < m; ++i)        {            scanf("%d %d %d",&u,&v,&cost);            add_edge(u,v,1,cost);        }        int res = min_cost_flow(0,n-1,2);        printf("Instance #%d: ",cnt);        if(res == -1) printf("Not possible\n");        else printf("%d\n",res);    }    return 0;}
阅读全文
0 0