hdu-2616

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 879    Accepted Submission(s): 639

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
  
  

   只想真心的说一句，深搜太难了，这个是我借鉴人家的代码的，其实我看得似懂非懂的，到现在还是有些许的不明白
  
  

   代码：
  
  

   //深搜
#include<iostream>
#include<string.h>
using namespace std;
int map[15][3];
int used[15];
int Min=100000000;
int type,sum;
int n;
void dfs(int k,int m){

   	if(k==n+1){

   		if(m>0)

   		   return;

   	}

   	if(m<=0){

   		type=1;

   		if(Min>sum)

   		   Min=sum;

   		return;

   	}

   	for(int i=1;i<=n;i++){

   		if(!used[i]){

   			int kk;

   		used[i]=1;

   		if(m<=map[i][2])

   		   kk=2*map[i][1];

   		else kk=map[i][1];

   		sum++;

   		m-=kk;

   		dfs(k+1,m);

   		used[i]=0;

   		sum--;

   		m+=kk;

   	}

   	}
}
int main(){

   	int m;

   	while(~scanf("%d %d",&n,&m)){

   		memset(map,0,sizeof(map));

   		memset(used,0,sizeof(used));

   		sum=0;

   		type=0;

   		Min=10000000;

   		for(int i=1;i<=n;i++)

   		   scanf("%d%d",&map[i][1],&map[i][2]);

   		dfs(1,m);

   		if(!type)

   			printf("-1\n");

   		else

   		    printf("%d\n",Min);

   	}

   	return 0;
} 

  
  

   一个小小的错误都会是这道题目通不过。。。。