Water Gate Management(二进制枚举组合排列模板 n个数的子集)

A dam has n water gates to let out water when necessary. Each water gate has its own capacity, water

path and affected areas in the downstream. The affected areas may have a risk of flood when the water

gate is open. The cost of potential damage caused by a water gate is measured in number calculated

from the number of people and areas estimated to get affected.

Suppose a water gate Gi has the volumetric flow rate of Fi m3/hour and the damage cost of Ci

. In

a certain situation, the dam has the volume V m3 of water to flush out within T hours. Your task is

to manage the opening of the water gates in order to get rid of at least the specified volume of water

within a limited time in condition that the damage cost is minimized.

For example, a dam has 4 water gates and their properties are shown in the following table.

Water Gate G1 G2 G3 G4

Flow rate (m3/hour) 720,000 50,000 130,000 1,200,000

Cost 120,000 60,000 50,000 150,000

Case 1: You have to flush out the water 5 million m3 within 7 hours. The minimum cost will be

120,000 by letting the water gate G1 open for 7 hours.

Case 2: You have to flush out the water 5 million m3 within 30 hours. The minimum cost will be

110,000 by letting the water gates G2 and G3 open, for example, G2 is open for 29 hours and G3 is

open for 28 hours.

Note that each water gate is independent and it can be open only in a unit of whole hour (no

fraction of hour).

Input

The first line includes an integer n indicating number of water gates (1 ≤ n ≤ 20). Then the next

n lines contain, in each line, two integers: Fi and Ci corresponding to the flow rate (m3/hour) and

the damage cost of the water gate Gi respectively. The next line contains the number m which is

the number of test cases (1 ≤ m ≤ 50). The following m lines contain, in each line, two integers: V

and T corresponding to the volume (m3

) of water to let out within T hours. (1 ≤ Fi

, V, Ci ≤ 109

,

1 ≤ T ≤ 1000)

Output

For each test case, print out the minimum cost in the exact format shown in the sample output below.

If it is not possible to let out the water of volume V in T hours from the dam, print out ‘IMPOSSIBLE’

(without quotation marks).

Sample Input

4

720000 120000

50000 60000

130000 50000

1200000 150000

3

5000000 7

5000000 30

63000000 24

Sample Output

Case 1: 120000

Case 2: 110000
Case 3: IMPOSSIBLE

思路:暴力+二进制枚举即可,只是涉及排列组合的内容用二进制

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
int c;
int f;
}no[50];
typedef long long ll;
ll cost[1500000];   //大于2的20次方
ll flow[1500000];
int main()
{
int n;
scanf("%d",&n);
for(int i=0;i<n;i++)
scanf("%d%d",&no[i].f,&no[i].c);
memset(cost,0,sizeof(cost));
memset(flow,0,sizeof(flow));

ll iend=1<<n;
for(int i=1;i<iend;i++)    //枚举2的n-1次方   例如: 1 2 3:的情况:1 2 3 12 13 23 123
for(int j=0;j<=n;j++)
if(i&1<<j)
{  cost[i]+=no[j].c;
flow[i]+=no[j].f;
}
int icase;
scanf("%d",&icase);
for(int i=1;i<=icase;i++)
{
int  v,t,flag=0;
ll ans=1000000000;
scanf("%d%d",&v,&t);
for(int j=1;j<=iend;j++)
{
if(flow[j]*t>=v)
{
flag=1;
ans=min(ans,cost[j]);
}
}
printf("Case %d: ",i);
if(flag)
printf("%lld\n",ans);
else
printf("IMPOSSIBLE\n");
}
return 0;
}