http://poj.org/problem?id=3308
r*c的地图
每一个大炮可以消灭一行一列的敌人
安装消灭第i行的大炮花费是ri
安装消灭第j行的大炮花费是ci
已知敌人坐标，同时消灭所有敌人，问最小花费
花费为所有大炮消费的乘积
乘转加：log（a*b*c）=log（a）+log（b）+log（c）
经典的最小点权覆盖
源点向行连，列向汇点连
第i行j列有敌人，点i向点j连inf边
最小点权覆盖=最小割
#include<cmath>
#include<queue>
#include<cstdio>
#include<cstring>
#include<algorithm> using namespace std; const double inf=;
const double eps=1e-; #define N 201
#define M 701 int src,decc; int tot;
int front[N],to[M<<],nxt[M<<];
double val[M<<]; int cur[N],lev[N]; queue<int>q; void add(int u,int v,double cap)
{
to[++tot]=v; nxt[tot]=front[u]; front[u]=tot; val[tot]=cap;
to[++tot]=u; nxt[tot]=front[v]; front[v]=tot; val[tot]=;
} bool bfs()
{
for(int i=;i<=decc;++i) lev[i]=-,cur[i]=front[i];
while(!q.empty()) q.pop();
lev[src]=;
q.push(src);
int now,t;
while(!q.empty())
{
now=q.front();
q.pop();
for(int i=front[now];i;i=nxt[i])
{
t=to[i];
if(lev[t]==- && val[i]>eps)
{
lev[t]=lev[now]+;
if(t==decc) return true;
q.push(t);
}
}
}
return false;
} double dinic(int now,double flow)
{
if(now==decc) return flow;
double rest=,delta;
int t;
for(int &i=cur[now];i;i=nxt[i])
{
t=to[i];
if(lev[t]>lev[now] && val[i]>eps)
{
delta=dinic(t,min(flow-rest,val[i]));
if(delta>eps)
{
rest+=delta;
val[i]-=delta; val[i^]+=delta;
if(fabs(rest-flow)<eps) break;
}
}
}
if(fabs(rest-flow)>eps) lev[now]=-;
return rest;
} int main()
{
int T;
scanf("%d",&T);
int n,m,k;
double x,ans;
int a,b;
while(T--)
{
scanf("%d%d%d",&n,&m,&k);
decc=n+m+;
tot=;
memset(front,,sizeof(front));
for(int i=;i<=n;++i)
{
scanf("%lf",&x);
add(src,i,log(x));
}
for(int i=;i<=m;++i)
{
scanf("%lf",&x);
add(i+n,decc,log(x));
}
for(int i=;i<=k;++i)
{
scanf("%d%d",&a,&b);
add(a,b+n,inf);
}
ans=;
while(bfs()) ans+=dinic(src,inf);
printf("%.4lf\n",exp(ans));
}
}
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8903 | Accepted: 2679 |
Description
It is year 2500 A.D. and there is a terrible war between the forces of the Earth and the Mars. Recently, the commanders of the Earth are informed by their spies that the invaders of Mars want to land some paratroopers in the m × n grid yard of one their main weapon factories in order to destroy it. In addition, the spies informed them the row and column of the places in the yard in which each paratrooper will land. Since the paratroopers are very strong and well-organized, even one of them, if survived, can complete the mission and destroy the whole factory. As a result, the defense force of the Earth must kill all of them simultaneously after their landing.
In order to accomplish this task, the defense force wants to utilize some of their most hi-tech laser guns. They can install a gun on a row (resp. column) and by firing this gun all paratroopers landed in this row (resp. column) will die. The cost of installing a gun in the ith row (resp. column) of the grid yard is ri (resp. ci ) and the total cost of constructing a system firing all guns simultaneously is equal to the product of their costs. Now, your team as a high rank defense group must select the guns that can kill all paratroopers and yield minimum total cost of constructing the firing system.
Input
Input begins with a number T showing the number of test cases and then, T test cases follow. Each test case begins with a line containing three integers 1 ≤ m ≤ 50 , 1 ≤ n ≤ 50 and 1 ≤ l ≤ 500 showing the number of rows and columns of the yard and the number of paratroopers respectively. After that, a line with m positive real numbers greater or equal to 1.0 comes where the ith number is ri and then, a line with n positive real numbers greater or equal to 1.0 comes where the ith number is ci. Finally, l lines come each containing the row and column of a paratrooper.
Output
For each test case, your program must output the minimum total cost of constructing the firing system rounded to four digits after the fraction point.
Sample Input
1
4 4 5
2.0 7.0 5.0 2.0
1.5 2.0 2.0 8.0
1 1
2 2
3 3
4 4
1 4
Sample Output
16.0000