poj3308

题意：一个矩阵，已知其中一些格会降落伞兵，每行每列都有一个武器，可以一次性消灭该行或该列的全部伞兵，每个武器对应不同的价格，若使用多个武器则总价是各个武器价钱的乘积，问消灭所有伞兵最少要多少钱。

分析：最小权覆盖集。难点在于武器总价不是加和而是乘积，那么我们需要把各个单价转化为以e为底的对数，这样再求对数加和的时候其实各个原单价之间是相乘的关系。建立二分图，每行对应一个节点，每列对应一个节点，各点权值等于其武器价钱的以e为底的对数，每个伞兵是连接其所在行列节点的一条边。对这个二分图求最小权独立集。转化为最小割，转化为最大流。

#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <cmath>
using namespace std;

#define inf 0x3f3f3f3f
#define maxn 55
#define maxl 505
#define N maxn * 2
#define E (N + maxl) * 2

int n, m, num;
int s, t;

struct edge
{
    int x, y, nxt;
    double c;
}bf[E];

int ne, head[N], cur[N], ps[N], dep[N];

void addedge1(int x, int y, double c)
{
    bf[ne].x = x;
    bf[ne].y = y;
    bf[ne].c = c;
    bf[ne].nxt = head[x];
    head[x] = ne++;
}

void addedge(int x, int y, double c)
{
    addedge1(x, y, c);
    addedge1(y, x, 0);
}

double flow(int n, int s, int t)
{
    double tr, res = 0;
    int i, j, k, f, r, top;
    while (1)
    {
        memset(dep, -1, n * sizeof(int));
        for (f = dep[ps[0] = s] = 0, r = 1; f !=r;)
            for (i = ps[f++], j = head[i]; j; j = bf[j].nxt)
            {
                if (bf[j].c && -1 == dep[k = bf[j].y])
                {
                    dep[k] = dep[i] + 1;
                    ps[r++] = k;
                    if (k == t)
                    {
                        f = r;
                        break;
                    }
                }
            }
        if (-1 == dep[t])
            break;

        memcpy(cur, head, n * sizeof(int));
        for (i = s, top = 0;;)
        {
            if (i == t)
            {
                for (k = 0, tr = inf; k  < top; ++k)
                    if (bf[ps[k]].c < tr)
                        tr = bf[ps[f = k]].c;
                for (k = 0; k < top; ++k)
                    bf[ps[k]].c -=tr, bf[ps[k]^1].c += tr;
                res += tr;
                i = bf[ps[top = f]].x;
            }
            for (j = cur[i]; cur[i]; j = cur[i] = bf[cur[i]].nxt)
                if (bf[j].c && dep[i] + 1 == dep[bf[j].y])
                    break;
            if (cur[i])
            {
                ps[top++] = cur[i];
                i = bf[cur[i]].y;
            }
            else
            {
                if (0 == top)
                    break;
                dep[i] = -1;
                i = bf[ps[--top]].x;
            }
        }
    }
    return res;
}

void input()
{
    scanf("%d%d%d", &n, &m, &num);
    s = 0;
    t = n + m + 1;
    for (int i = 1; i <= n; i++)
    {
        double a;
        scanf("%lf", &a);
        addedge(s, i, log(a));
    }
    for (int i = n + 1; i <= n + m; i++)
    {
        double a;
        scanf("%lf", &a);
        addedge(i, t, log(a));
    }
    for (int i = 0; i < num; i++)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        addedge(a, n + b, inf);
    }
}

int main()
{
    //freopen("t.txt", "r", stdin);
    int case_num;
    scanf("%d", &case_num);
    while (case_num--)
    {
        ne = 2;
        memset(head, 0, sizeof(head));
        input();
        printf("%.4f\n", exp(flow(n + m + 2, s, t)));
    }
    return 0;
}