poj2516

/***************************************************************\
* Author: Hu Wenbiao
* Created Time: Sat 02 Oct 2010 08:31:09 PM CST
* File Name: main.cpp
* Description: 最小费用最大流。对每种货物用一次mcmf。
\***************************************************************/
//*========================*Head File*========================*\\

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<queue>
/*----------------------*Global Variable*----------------------*/

#define maxn 105
#define inf 0x3f3f3f3f
struct edge {
    int v, w, f, c, next;
} e[5100];
int vst[maxn], dis[maxn], start[maxn], p[maxn];
int tot, N, M, K, flow_sum, ans, need[60][60], supply[60][60], cost,
    good_sum;

//*=======================*Main Program*=======================*//
using namespace std;

void _add(int v, int w, int f, int c)
{
    e[tot].v = v;
    e[tot].w = w;
    e[tot].f = f;
    e[tot].c = c;
    e[tot].next = start[v];
    start[v] = tot++;
}

void add(int v, int w, int f, int c)
{
    _add(v, w, f, c);
    _add(w, v, 0, -c);
}

bool spfa(int s, int t, int n)
{
    int v, w;
    queue < int >q;
    for (int i = 0; i < n; i++) {
	p[i] = -1;;
	vst[i] = 0;
	dis[i] = inf;
    }
    vst[s] = 1;
    dis[s] = 0;
    q.push(s);

    while (!q.empty()) {
	v = q.front();
	q.pop();
	vst[v] = false;
	for (int i = start[v]; i != -1; i = e[i].next) {
	    if (e[i].f) {
		w = e[i].w;
		if (dis[w] > dis[v] + e[i].c) {
		    dis[w] = dis[v] + e[i].c;
		    p[w] = i;
		    if (!vst[w]) {
			vst[w] = true;
			q.push(w);
		    }
		}
	    }
	}
    }
    return dis[t] != inf;
}

int mcmf(int s, int t, int n)
{
    int ans = 0, flow = inf, i;
    while (spfa(s, t, n)) {
	for (i = p[t]; i != -1; i = p[e[i].v])
	    if (e[i].f < flow)
		flow = e[i].f;
	for (i = p[t]; i != -1; i = p[e[i].v]) {
	    e[i].f -= flow;
	    e[i ^ 1].f += flow;
	}
	ans += dis[t] * flow;
	flow_sum += flow;
    }
    return ans;
}

int main()
{
//freopen("input","r",stdin);
    while (scanf("%d%d%d", &N, &M, &K) == 3 && N && M && K) {
	ans = 0;
	for (int i = 1; i <= N; i++) {
	    for (int j = 1; j <= K; j++) {
		scanf("%d", &need[i][j]);
	    }
	}
	for (int i = 1; i <= M; i++) {
	    for (int j = 1; j <= K; j++) {
		scanf("%d", &supply[i][j]);
	    }
	}
	bool flag = true;
	for (int k = 1; k <= K; k++) {	//第k种货物
	    flow_sum = 0;
	    tot = 0;
	    memset(start, -1, sizeof(start));
	    for (int i = 1; i <= N; i++) {
		for (int j = 1; j <= M; j++) {
		    scanf("%d", &cost);
		    if (flag)
			add(j, M + i, 10000000, cost);
		}
	    }
	    for (int i = 1; i <= M; i++) {
		if (flag)
		    add(0, i, supply[i][k], 0);
	    }
	    for (int i = 1; i <= N; i++) {
		if (flag)
		    add(M + i, M + N + 1, need[i][k], 0);
	    }
	    if (flag) {
		ans += mcmf(0, N + M + 1, N + M + 2);

		good_sum = 0;
		for (int t = 1; t <= N; t++)
		    good_sum += need[t][k];

		if (flow_sum < good_sum)	//所能达到的最大流小于需要的
		    flag = false;
	    }
	}
	if (flag)
	    printf("%d\n", ans);
	else
	    printf("-1\n");
    }
}