UVA1486 Transportation(最小费用最大流)

题意:

N个点(1<=N<=100)M(1<=M<=5000)条有向边,运送K(1<=K<=100)个单位商品,每条边都有一个系数ai(0<ai<=100)和容量ci(ci<=5),若运送x个单位的话,就得交ai*x^2美元.

问你,将K个单位商品从1运到N,最小花费.若运送不到,输出-1;

分析:因为容量Ci比较小,故可以拆边,把容量为Ci的边拆成Ci条边,每条边的花费为x_i*x_i*ai-(x_i-1)(x_i-1)*ai;

详细见代码

// File Name: 1486.cpp
// Author: Zlbing
// Created Time: 2013/4/26 13:53:55

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
using namespace std;
#define CL(x,v); memset(x,v,sizeof(x));
#define INF 0x3f3f3f3f
#define LL long long
#define REP(i,r,n) for(int i=r;i<=n;i++)
#define RREP(i,n,r) for(int i=n;i>=r;i--)
const int MAXN=205;
struct Edge{
    int from,to,cap,flow,cost;
};
struct MCMF{
    int n,m,s,t;
    vector<Edge>edges;
    vector<int> G[MAXN];
    int inq[MAXN];
    int d[MAXN];
    int p[MAXN];
    int a[MAXN];
    void init(int n){
        this->n=n;
        for(int i=0;i<=n;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});
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }
    bool BellmanFord(int s,int t,int& flow,int& cost){
        for(int i=0;i<=n;i++)d[i]=INF;
            CL(inq,0);
        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+=d[t]*a[t];
        int u=t;
        while(u!=s){
            edges[p[u]].flow+=a[t];
            edges[p[u]^1].flow-=a[t];
            u=edges[p[u]].from;
        }
        return true;
    }
    int Mincost(int s,int t,int k){
        int flow=0,cost=0;
        while(BellmanFord(s,t,flow,cost));
        if(flow!=k)
            return -1;
        else
        return cost;
    }
};

int n,m,k;
MCMF solver;
int main()
{
    while(~scanf("%d%d%d",&n,&m,&k))
    {
        int a,b,c,d;
        solver.init(n);
        REP(i,1,m)
        {
            scanf("%d%d%d%d",&a,&b,&c,&d);
            REP(j,1,d)
            {
                solver.AddEdge(a,b,1,j*j*c-(j-1)*(j-1)*c);
            }
        }
        solver.AddEdge(0,1,k,0);
        int s=0,t=n;
        int ans=solver.Mincost(s,t,k);
        printf("%d\n",ans);
    }
    return 0;
}