POJ 2516

题意:

• n个店主,m个供货商(仓库),k种商品
• 给出n个店主的k种商品的需求,m个仓库存放k种商品的个数,和对于每种商品$k_i$来说从$m_i$运送到$n_i$的运费
• 求最小花费, 如果商品不足则输出-1

思路:

1. 分k次计算：每个商品分别求最小花费,然后加和，建立二分图后首先判断商品数量是否足够给店主.**然后裸KM()**

存数据+建图

1. shop[i][k]存第i个店主的第$k_i$种商品的需求个数，sup[i][k]存第i个仓库存的第$k_i$中商品的个数.cost[t][i][j]存商品t从仓库j运到店主i的花费
   • KM一般是左部分少右部分多的完备匹配
2. 对于第$k_i$种商品: 将店主需要的所有该种商品的个数拆成二分图的左部分的点,仓库拥有的所有该商品拆成二分图的右部分的点,
3. 边换成负值套用KM求最大带权匹配

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn = 55;
const int inf = -0x3f3f3f3f;
int shop[maxn][maxn],sup[maxn][maxn],cost[maxn][maxn][maxn];

int n,m,k,cntA,cntB,delta;
int belongA[maxn*3],belongB[maxn*3];
int w[maxn*3][maxn*3],la[maxn*3],lb[maxn*3],match[maxn*3];
bool va[maxn*3],vb[maxn*3];

void read() {
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=k; ++j)
            scanf("%d",&shop[i][j]);
    for (int i=1; i<=m; ++i)
        for (int j=1; j<=k; ++j)
            scanf("%d",&sup[i][j]);
    for (int t=1; t<=k; ++t)
        for (int i=1; i<=n; ++i)
            for (int j=1; j<=m; ++j)
                scanf("%d",&cost[t][i][j]);
}

int dfs(int u) {
    va[u] = 1;
    for (int i=1; i<=cntB; ++i) {
        if (!vb[i]) {
            if (la[u]+lb[i] - w[u][i] == 0) {
                vb[i] = 1;
                if (!match[i] || dfs(match[i])) {
                    match[i] = u;
                    return 1;
                }
            } else delta = min(delta,la[u]+lb[i]-w[u][i]);
        }
    }
    return 0;
}


int KM () {
    for (int i=0; i<maxn*3; ++i) la[i] = -inf;
    memset(lb,0,sizeof lb);
    for (int i=1; i<=cntA; ++i)
        for (int j=1; j<=cntB; ++j)
            la[i] = max(la[i],w[i][j]);
    memset(match,0,sizeof match);
    for (int i=1; i<=cntA; ++i) {
        while (1) {
            memset(va,0,sizeof va);
            memset(vb,0,sizeof vb);
            delta = inf;
            if (dfs(i)) break;
            for (int i=1; i<=cntB; ++i) {
                if (va[i]) la[i] -= delta;
                if (vb[i]) lb[i] += delta;
            }
        }
    }
    int ans = 0;
    for (int i=1; i<=cntB; ++i)
        if (match[i]) ans += w[match[i]][i];
    return ans;
}
int main() {
    freopen("1.in","r",stdin);
    while (scanf("%d%d%d",&n,&m,&k) != EOF && (n+m+k)) {
        read();
        // 判断仓库的货物是否足够
        bool flag = 1;
        for (int i=1; i<=k; ++i) {
            int need =0, have = 0;
            for (int j=1; j<=n; ++j)
                need += shop[j][i];
            for (int j=1; j<=m; ++j)
                have += sup[j][i];
            if (need > have) {
                flag = 0;
                printf("-1\n");
                break;
            }
        }
        if (!flag) continue;
        int ans = 0;
        // KM分别处理每个货物
        for (int t=1; t<=k; ++t) {
            cntA = cntB = 0;
            for (int i=1; i<=n; ++i)
                for (int j=1; j<=shop[i][t]; ++j)
                    belongA[++cntA] = i; //给同一种货物的所有货物标号顺便 标记属于哪个店家
            for (int i=1; i<=m; ++i)
                for (int j=1; j<=sup[i][t]; ++j)
                    belongB[++cntB] = i; //存储点的货物属于哪个仓库,并且给每个相同货物标记号码
            for (int i=1; i<=cntA; ++i)
                for (int j=1; j<=cntB; ++j)
                    w[i][j] = -cost[t][belongA[i]][belongB[j]];
            ans += KM();
        }
        printf("%d\n",-ans);
    }
    return 0;
}