POJ 2516 Minimum Cost 费用流

题目源：https://vjudge.net/problem/POJ-2516

             http://poj.org/problem?id=2516

模板源：https://blog.youkuaiyun.com/Adolphrocs/article/details/84558274

题解：费用流 求出对于每个k，supply 与 shopkeeper的网络流，然后累加

#include<cstdio>
#include<string.h>
#include<queue>
#include<algorithm>
#include <math.h>
#define inf 1e8
using namespace std;
const int MAXN = 200;
const int MAXM = 51000;
const double eps = 1e-7;
const float EXP = exp(1.0);

int num, head[MAXN];
struct Edge {
    int u, v, flow, nxt;
    double cost;
    Edge() {}
    Edge(int u, int v, int flow, double cost, int nxt): u(u), v(v), flow(flow), cost(cost), nxt(nxt) {}
} E[MAXM];

int n, m;

void init() {
    num = 0;
    memset(head, -1, sizeof head);
}

void add(int u, int v, int flow, double cost) {
    E[num] = Edge(u, v, flow, cost, head[u]);
    head[u] = num++;
    E[num] = Edge(v, u, 0, -cost, head[v]);
    head[v] = num++;
}
int pre[MAXN], mi[MAXN];
double dis[MAXN];
bool vis[MAXN];
int s, t;
double minCost;
int flow;
int que[MAXN];
bool SPFA() {
    for(int i = 0; i <= t; i++)
    	vis[i] = 0, dis[i] = inf, pre[i] = -1;
    dis[s] = 0, mi[s] = inf, vis[s] = true;
    int l = 0, r = 1;
    que[l] = s;
    while(l != r) {
        int u = que[l++];
        if(l == MAXN) l = 0;
        vis[u] = 0;
        for(int i = head[u]; i + 1; i = E[i].nxt) {
            int v = E[i].v;
            if(E[i].flow > 0 && dis[v] > dis[u] + E[i].cost+eps) {
                dis[v] = dis[u] + E[i].cost;
                pre[v] = i;
                mi[v] = min(mi[u], E[i].flow);
                if(!vis[v]) {
                    vis[v] = 1;
                    que[r++] = v;
                    if(r >= MAXN) r -= MAXN;
                }
            }
        }
    }
    if(pre[t] == -1) return false;
    flow += mi[t];
    minCost += mi[t] * dis[t];
    int u = t;
    for(int i = pre[u]; i + 1; i = pre[E[i].u]) {
        E[i].flow -= mi[t];
        E[i ^ 1].flow += mi[t];
    }
    return true;
}

double Mincost() {
    minCost = 0, flow = 0;
    while(SPFA());
    return minCost;
}

int main() {
    int N, M, K;
    int board[55];
    int need[55][55];
    int have[55][55];
    int price[55][55][55];//K*N*M;
    while(scanf("%d%d%d", &N, &M, &K) && (N || M || K)){
        memset(board, 0, sizeof board);
        for(int i = 1; i <= N; ++i){
            for(int j = 1; j <= K; ++j){
                scanf("%d", &need[i][j]);
                board[j] -= need[i][j];
            }
        }
        for(int i = 1; i <= M; ++i){
            for(int j = 1; j <= K ; ++j){
                scanf("%d", &have[i][j]);
                board[j] += have[i][j];
            }
        }
        for(int i = 1; i <= K; ++i){
            for(int j = 1; j <= N; ++j){
                for(int k = 1; k <= M; ++k){
                    scanf("%d", &price[i][j][k]);
                }
            }
        }
        bool flag = false;
        for(int i=1 ; i<=K ; ++i){
            if(board[i] < 0){
                printf("-1\n");
                flag = true;
                break;
            }
        }
        if(flag)continue;
        int sum = 0;
        for(int i = 1 ; i <= K ; ++i){
            init();
            for(int j = 1; j <= N; ++j) add(j + 50, MAXN - 1, need[j][i], 0);//建立 N到超汇的边
            for(int j = 1; j <= M; ++j) add(0 , j, have[j][i], 0);//建立超源到 M的边
            for(int j = 1; j <= N; ++j){//建立 M到N的边
                for(int k = 1 ;k <= M ; ++k){
                    add(k, j + 50, inf, price[i][j][k]);
                }
            }
            s = 0; t = MAXN - 1;
            sum += Mincost();
        }
        printf("%d\n",sum);
    }
    return 0;
}