hdu 2616 Kill the monster(stl全排列+暴力)

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 502    Accepted Submission(s): 345

Problem Description

There is a mountain near yifenfei’s hometown. On the mountain lived a big monster. As a hero in hometown, yifenfei wants to kill it.
Now we know yifenfei have n spells, and the monster have m HP, when HP <= 0 meaning monster be killed. Yifenfei’s spells have different effect if used in different time. now tell you each spells’s effects , expressed (A ,M). A show the spell can cost A HP to monster in the common time. M show that when the monster’s HP <= M, using this spell can get double effect.

 

 

Input

The input contains multiple test cases.
Each test case include, first two integers n, m (2<n<10, 1<m<10^7), express how many spells yifenfei has.
Next n line , each line express one spell. (Ai, Mi).(0<Ai,Mi<=m).

 

 

Output

For each test case output one integer that how many spells yifenfei should use at least. If yifenfei can not kill the monster output -1.

 

 

Sample Input

 3 100

10 20

45 89

5 40

 

3 100

10 20

45 90

5 40

 

3 100

10 20

45 84

5 40

 

Sample Output

3 2 -1

分析：

（1）因为数据比较少，每一组最多有是个数据，可以用全排列暴力计算找出最小的组合

（2）这里借助于stl算法库中的next_permutation()全排列算法。暴力把所有的排列方法计算一遍。然后找出最小的组合方法。

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
int main()
{
    int o[10],a[10],m[10];
    int n,M,count,min;

    while(cin>>n>>M)
    {
        for(int i=0;i<n;i++)
        cin>>a[i]>>m[i];
        for(int i=0;i<n;i++)
        o[i] = i;
        min = 100;
        do{
            count = 0;
            int tem = M;
            for(int i=0;i<n;i++)
            {
                count++;
                if(tem<=m[o[i]])
                tem -= a[o[i]]*2;
                else tem -= a[o[i]];
                if(tem<=0)
                {
                    if(count<min)
                    min = count;
                    break;
                }
            }
        }while(next_permutation(o,o+n));
        if(min==100)printf("-1\n");
        else printf("%d\n",min);
    }

    return 0;
}