Hoj 2543 Stone IV

题目链接：http://acm.hit.edu.cn/hoj/problem/view?id=2543

本题练习最小费用流。

用队列优化的Bellmanford来寻找增广路，每次找最小费用可行流，对于一条边(u,v)，我们规定cap[v][u] = 0,cost[v][u] = -cost[u][v].

由于是无向图，加上反向边，再加上我们要拆成两条边来处理本题，边数×8，即每一条边Add八次类型。如果要处理具有平行边和反向边的情况，如本题，不能采用临界矩阵，本题用前向星。

#include <iostream>#include <stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#include <vector>#include <queue>#include <algorithm>using namespace std;#define INF 0x3f3f3f3f#define Maxn 1005#define Maxm 80005int uE[Maxm];int vE[Maxm];int cap[Maxm];int flow[Maxm];int cost[Maxm];int next[Maxm];int first[Maxn];int pre[Maxn];int dist[Maxn];//最小花费int inq[Maxn];int a[Maxn];//残留网络int e;int n,m,c,p;void init(){    memset(first,-1,sizeof(first));    memset(next,-1,sizeof(next));}void addEdge(int x,int y,int a,int b){    uE[e] = x,vE[e] = y,cap[e] = a,cost[e] = b;    next[e] = first[x];    first[x] = e;    e++;}int BellmanFord_EdmondsKarp(int s,int t){    int f = 0;    memset(flow,0,sizeof(flow));    queue<int> q;    for(;;)    {        memset(inq,0,sizeof(inq));        memset(a,0,sizeof(a));        memset(dist,0x3f,sizeof(dist));        q.push(s);        inq[s] = 1,dist[s] = 0,a[s] = INF,pre[s] = -1;        while(!q.empty())        {            int u = q.front();            q.pop();            inq[u] = 0;            for(int i=first[u]; i!=-1; i=next[i])            {                int v = vE[i];                if(flow[i]<cap[i] && dist[u] + cost[i]<dist[v])                {                    a[v] = a[u] < cap[i] - flow[i] ? a[u] : cap[i] - flow[i];                    dist[v] = dist[u] + cost[i];                    pre[v] = i;//注意此时pre[]的意义是某点的前向边                    if(!inq[v])                    {                        inq[v] = 1,q.push(v);                    }                }            }        }        if(dist[t] == INF || a[t] == 0) return f;        //注意此处转换成long long ,防止数据超出范围,得出相反结论        if((long long)a[t] * dist[t]<=(long long)c)        {            for(int i=pre[t];i!=-1;i=pre[uE[i]])            {                flow[i] += a[t];                flow[i^1] -= a[t];            }            f += a[t];            c -= a[t]*dist[t];        }        else        {            return f + c/dist[t];        }    }    return 0;}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt","r",stdin);#endif    int T;    int x,y,c1,c2;    int s,t;    scanf(" %d",&T);    while(T--)    {        init();        scanf(" %d %d %d %d",&n,&m,&c,&p);        e = 0,s = n,t = 1;        addEdge(s,0,INF,p),addEdge(0,s,0,-p);        for(int i=0; i<m; i++)        {            scanf(" %d %d %d %d",&x,&y,&c1,&c2);            addEdge(x,y,c1,0),addEdge(y,x,0,0),addEdge(x,y,INF,c2),addEdge(y,x,0,-c2);            addEdge(y,x,c1,0),addEdge(x,y,0,0),addEdge(y,x,INF,c2),addEdge(x,y,0,-c2);        }        int ans = BellmanFord_EdmondsKarp(s,t);        printf("%d\n",ans);    }    return 0;}