题意

敌人侵略r*c的地图。为了消灭敌人，可以在某一行或者某一列安置超级大炮。每一个大炮可以瞬间消灭这一行（或者列）的敌人。安装消灭第i行的大炮消费是ri。安装消灭第j行的大炮消费是ci现在有n个敌人，告诉你这n个敌人的坐标，让你同时消灭这些敌人，为你最小花费是多少。花费的定义：每个大炮消费的乘积。

思路

非常经典的最小点权覆盖集问题，同最大流建模就可以了，建模方法可见胡伯涛论文《最小割模型在信息学竞赛中的应用》。

这道题的模型转换成最小点权覆盖集的方法可见这里.

这里的点权最大是乘积的，只要对权值取对数就把乘法转换成加法了~~

代码

[cpp]
#include <iostream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <stack>
#include <queue>
#include <algorithm>
#include <string>
#include <cstring>
#define MID(x,y) ((x+y)/2)
#define MEM(a,b) memset(a,b,sizeof(a))
#define REP(i, begin, m) for (int i = begin; i < begin+m; ++ i)
using namespace std;
const int MAXV = 1005;
const int MAXE = 1005;
const int oo = 0x3fffffff;

template
struct Dinic{
struct flow_node{
int u, v;
T flow;
int opp;
int next;
}arc[2*MAXE];
int vn, en, head[MAXV];
int cur[MAXV];
int q[MAXV];
int path[2*MAXE], top;
int dep[MAXV];
void init(int n){
vn = n;
en = 0;
MEM(head, -1);
}
void insert_flow(int u, int v, T flow){
arc[en].u = u;
arc[en].v = v;
arc[en].flow = flow;
arc[en].next = head[u];
head[u] = en ++;

arc[en].u = v;
arc[en].v = u;
arc[en].flow = 0;
arc[en].next = head[v];
head[v] = en ++;
}
bool bfs(int s, int t){
MEM(dep, -1);
int lq = 0, rq = 1;
dep[s] = 0;
q[lq] = s;
while(lq < rq){ int u = q[lq ++]; if (u == t){ return true; } for (int i = head[u]; i != -1; i = arc[i].next){ int v = arc[i].v; if (dep[v] == -1 && arc[i].flow > 0){
dep[v] = dep[u] + 1;
q[rq ++] = v;
}
}
}
return false;
}
T solve(int s, int t){
T maxflow = 0;
while(bfs(s, t)){
int i, j;
for (i = 1; i <= vn; i ++) cur[i] = head[i];
for (i = s, top = 0;;){
if (i == t){
int mink;
T minflow = 0x7fffffff; //要比容量的oo大
for (int k = 0; k < top; k ++) if (minflow > arc[path[k]].flow){
minflow = arc[path[k]].flow;
mink = k;
}
for (int k = 0; k < top; k ++)
arc[path[k]].flow -= minflow, arc[path[k]^1].flow += minflow;
maxflow += minflow;
top = mink;
i = arc[path[top]].u;
}
for (j = cur[i]; j != -1; cur[i] = j = arc[j].next){
int v = arc[j].v;
if (arc[j].flow && dep[v] == dep[i] + 1)
break;
}
if (j != -1){
path[top ++] = j;
i = arc[j].v;
}
else{
if (top == 0) break;
dep[i] = -1;
i = arc[path[-- top]].u;
}
}
}
return maxflow;
}
};
Dinic dinic;
double r[55], c[55];

int main(){
//freopen("test.in", "r", stdin);
//freopen("test.out", "w", stdout);
int n, m, l;
int t;
scanf("%d", &t);
while(t --){
scanf("%d %d %d", &n, &m, &l);
dinic.init(n+m+2);
REP(i, 1, n){
scanf("%lf", &r[i]);
dinic.insert_flow(n+m+1, i, log(r[i]));
}
REP(i, 1, m){
scanf("%lf", &c[i]);
dinic.insert_flow(i+n, n+m+2, log(c[i]));
}

REP(i, 1, l){
int u, v;
scanf("%d %d", &u, &v);
dinic.insert_flow(u, v+n, oo);
}
printf("%.4f\n", exp(dinic.solve(n+m+1, n+m+2)));
}
return 0;
}
[/cpp]