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
大概就是n表示的三次魔法，m表示的是hp值然后，下面三个是每个法术的普通伤害和低于他规定的值后翻倍；
用深度优先搜索暴力解题，其实dfs本质还是跟递归有关，还是不能学的僵化了。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
using namespace std;
int used[100];//标志变量
int a[100],b[100];//存储两个数
int ans;//判断变量
int cut=0;
int n, m;
void dfs(int bllod,int cut)
{
if(cut>n)
return ;
if(bllod<=0&&cut<ans)
{
ans=cut;
return ;
}
for(int i=0;i<n;i++){
if(!used[i])
{
used[i]=1;
if(bllod<=b[i])
dfs(bllod-2*a[i],cut+1);
else
dfs(bllod-a[i],cut+1);
used[i]=0;
}
}
}
int main(){
while(cin>>n>>m)
{
for(int i=0;i<n;i++)
{
cin >> a[i] >>b[i];
}
memset(used,0,sizeof(used));
ans=1314;
dfs(m,cut);
if(ans!=1314)
cout<<ans<<endl;
else cout<<-1<<endl;
cut=0;
}
return 0;
}
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