poj3422 来一道费用流

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题目链接 

poj3422

真是感觉网络流越来越神奇了。

1、最小费用流 -- 最大费用流之间转化

2、拆点

3、费用 OR 流量  (cost, flow)

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>

using namespace std;

#define clr(x) memset(x,0,sizeof(x))
#define fp1 freopen("in.txt","r",stdin)
#define fp2 freopen("out.txt","w",stdout)
#define pb push_back

#define INF 0x3c3c3c3c
typedef long long LL;

//网络中可以有负权边，但不能有负权圈。
//每次先调用init(); 初始maxn、INF之类。
//Addedge(cap, flow):cap是容量,flow是流量。
const int maxn = 100000;
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];      //是否在队列中
    int d[maxn];        //Bellman-Ford
    int p[maxn];        //上一条弧
    int a[maxn];        //可改进量

    void init(int n)    //感觉每次调用都要调用init初始化哪
    {
        this->n = n;
        for(int i = 0;i < n;i++) G[i].clear();
        edges.clear();
        //clr(p); clr(a);  clr(d);
    }

    void addedge(int from, int to, int cost, int cap)
    {
        edges.pb((Edge){from, to, cap, 0, cost});
        edges.pb((Edge){to, from, 0, 0, -cost});
        m = edges.size();
        G[from].pb(m-2);
        G[to].pb(m-1);
    }

    //其他成员函数

    bool spfa(int s, int t, int &flow, int &cost)
    {
        for(int i = 0;i < n;i++) d[i] = INF;
        clr(inq);
        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;  //s-t不连通，失败退出
        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 flow = 0, cost = 0;
        while(spfa(s, t, flow, cost));
        return cost;
    }

};
MCMF flow;
int map1[55][55];
int main()
{
    //freopen("in.txt", "r", stdin);
    int n, m;
    while (scanf("%d %d", &n, &m) == 2) {

        flow.init(n*n*2+100);
		int i, j;

		for (int i=1; i <= n;i++) {
			for (int j=1; j <= n;j++) {
				scanf("%d", &map1[i][j]);
			}
		}

		for (i = 1; i <= n; i++) {
			for (j = 1; j <= n; j++) {
				int u = (i - 1) * n + j;
				flow.addedge(u, n*n+u, -map1[i][j], 1); //reversed
                flow.addedge(u, n*n+u, 0, m-1);

				if(j != n){
                    flow.addedge(u, u+1, 0, m);
                    flow.addedge(n*n+u, u+1, 0, m);
				}
				if(i != n){
                    flow.addedge(u, u+n, 0, m);
                    flow.addedge(n*n+u, u+n, 0, m);
				}
			}
		}
		flow.addedge(0, 1, 0, m);                 //SuperSource
		flow.addedge(n*n*2, n * n * 2 + 1, 0, m); //SuperSink

        cout<<-flow.Mincost(0, n*n*2+1);
	}
    return 0;
}

poj3422 来一道费用流

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原文地址：http://blog.csdn.net/cgf1993/article/details/38150965