Kill the monster（深搜）

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

Author yifenfei

Source 奋斗的年代

Recommend yifenfei

题意：

山上有个大boss，让你去击杀，boss有m多的血，而你有n种技能，当每种技能在使用时boss的血量低于某个值时，伤害翻倍！问你使用的最小技能数是多少！

思路：

深搜，搜出每种情况，然后比较，得出最小的那个！

代码：

#include <bits/stdc++.h>
using namespace std;
int n,m;
struct node
{
	int x;
	int y;
}a[11];
bool vis[11];
int sum;
void dfs(int aa,int b)
{
    int i,j; 
	if(b>=sum)     //剪枝，要学会
	return;
	if(aa<=0)
	{
		if(sum>b)
		sum=b;
		return;
	}
	for (i=0;i<n;i++)
	{
		if(!vis[i])
		{
			vis[i]=1;
			if(aa<=a[i].y)
			dfs(aa-2*a[i].x,b+1);
		    else
		   	dfs(aa-a[i].x,b+1);
		    vis[i]=0;
		}
	}
	
}
int main()
{
	int i,j;
	while(cin>>n>>m)
	{
		
		for(i=0;i<n;i++)
		cin>>a[i].x>>a[i].y;
		memset(vis,0,sizeof(vis));
		sum=11;
		dfs(m,0);
		if(sum==11)
		cout<<"-1"<<endl;
		else
		cout<<sum<<endl; 
		
	}
	return 0;
}