本文主要是介绍POJ 3422 最小费用最大流 zkw或者普通版本，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
建图的话
每个点拆成两个点u, u'，连一条容量为1费用为金币数的边，再连一条容量为k，费用为0的边
然后每个点和他右边或者下边的点连边 i'->j这样连
然后源点连1点，右下角那个点去连汇点，容量都为k，费用为0
普通写法
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 5555
#define MAXM 55555
#define INF 100000007
using namespace std;
struct EDGE
{int v, cap, cost, next, re; // re记录逆边的下标。
} edge[MAXM];
int n, m, ans, flow, src, des;
int e, head[MAXN];
int que[MAXN], pre[MAXN], dis[MAXN];
bool vis[MAXN];
void init()
{e = ans = flow = 0;memset(head, -1, sizeof(head));
}
void add(int u, int v, int cap, int cost)
{edge[e].v = v;edge[e].cap = cap;edge[e].cost = cost;edge[e].next = head[u];edge[e].re = e + 1;head[u] = e++;edge[e].v = u;edge[e].cap = 0;edge[e].cost = -cost;edge[e].next = head[v];edge[e].re = e - 1;head[v] = e++;
}
bool spfa()
{int i, h = 0, t = 1;for(i = 0; i <= n; i ++){dis[i] = INF;vis[i] = false;}dis[src] = 0;que[0] = src;vis[src] = true;while(t != h){int u = que[h++];h %= n;vis[u] = false;for(i = head[u]; i != -1; i = edge[i].next){int v = edge[i].v;if(edge[i].cap && dis[v] > dis[u] + edge[i].cost){dis[v] = dis[u] + edge[i].cost;pre[v] = i;if(!vis[v]){vis[v] = true;que[t++] = v;t %= n;}}}}if(dis[des] == INF) return false;return true;
}
void end()
{int u, p, mi = INF;for(u = des; u != src; u = edge[edge[p].re].v){p = pre[u];mi = min(mi, edge[p].cap);}for(u = des; u != src; u = edge[edge[p].re].v){p = pre[u];edge[p].cap -= mi;edge[edge[p].re].cap += mi;ans += mi * edge[p].cost; // cost记录的为单位流量费用，必须得乘以流量。}flow += mi;
}
int nt, k;
void build()
{init();int w;for(int i = 1; i <= nt; i++)for(int j = 1; j <= nt; j++){scanf("%d", &w);int id = (i - 1) * nt + j;add(id, id + nt * nt, 1, -w);add(id, id + nt * nt, k, 0);if(i < nt) add(id + nt * nt, id + nt, k, 0);if(j < nt) add(id + nt * nt, id + 1, k, 0);}src = nt * nt * 2 + 1;des = nt * nt * 2 + 2;n = des;add(src, 1, k, 0);add(nt * nt * 2, des, k, 0);
}
void MCMF()
{init();build();while(spfa()) end();
}
int main()
{while(scanf("%d%d", &nt, &k) != EOF){MCMF();printf("%d\n", -ans);}return 0;
}
zkw写法
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdio>
#include <cmath>
#include <queue>
#include <deque>
#include <map>
#include <set>
#define eps 1e-5
#define MAXN 5555
#define MAXM 55555
#define INF 100000007
using namespace std;
struct EDGE
{int cost, cap, v;int next, re;
}edge[MAXM];
int head[MAXN], e;
int vis[MAXN], d[MAXN];
int ans, cost, src, des, n;
void init()
{memset(head, -1, sizeof(head));e = 0;ans = cost = 0;
}
void add(int u, int v, int cap, int cost)
{edge[e].v = v;edge[e].cap = cap;edge[e].cost = cost;edge[e].re = e + 1;edge[e].next = head[u];head[u] = e++;edge[e].v = u;edge[e].cap = 0;edge[e].cost = -cost;edge[e].re = e - 1;edge[e].next = head[v];head[v] = e++;
}
int aug(int u, int f)
{if(u == des){ans += cost * f;return f;}vis[u] = 1;int tmp = f;for(int i = head[u]; i != -1; i = edge[i].next)if(edge[i].cap && !edge[i].cost && !vis[edge[i].v]){int delta = aug(edge[i].v, tmp < edge[i].cap ? tmp : edge[i].cap);edge[i].cap -= delta;edge[edge[i].re].cap += delta;tmp -= delta;if(!tmp) return f;}return f - tmp;
}
bool modlabel()
{for(int i = 0; i <= n; i++) d[i] = INF;d[des] = 0;deque<int>Q;Q.push_back(des);while(!Q.empty()){int u = Q.front(), tmp;Q.pop_front();for(int i = head[u]; i != -1; i = edge[i].next)if(edge[edge[i].re].cap && (tmp = d[u] - edge[i].cost) < d[edge[i].v])(d[edge[i].v] = tmp) <= d[Q.empty() ? src : Q.front()] ? Q.push_front(edge[i].v) : Q.push_back(edge[i].v);}for(int u = 1; u <= n; u++)for(int i = head[u]; i != -1; i = edge[i].next)edge[i].cost += d[edge[i].v] - d[u];cost += d[src];return d[src] < INF;
}
void costflow()
{while(modlabel()){do{memset(vis, 0, sizeof(vis));}while(aug(src, INF));}
}
int nt, k;
int main()
{int w;while(scanf("%d%d", &nt, &k) != EOF){init();for(int i = 1; i <= nt; i++)for(int j = 1; j <= nt; j++){scanf("%d", &w);int id = (i - 1) * nt + j;add(id, id + nt * nt, 1, -w);add(id, id + nt * nt, k, 0);if(i < nt) add(id + nt * nt, id + nt, k, 0);if(j < nt) add(id + nt * nt, id + 1, k, 0);}src = nt * nt * 2 + 1;des = nt * nt * 2 + 2;n = des;add(src, 1, k, 0);add(nt * nt * 2, des, k, 0);costflow();printf("%d\n", -ans);}return 0;
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