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");

    }

}