HDU 2616 Kill the monster

Kill the monster

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 718    Accepted Submission(s): 511

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

奮鬥的年代

 

 

Recommend

yifenfei

 

思路：簡單DFS便可

 

代碼：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int minn,sum,flag,n,m,ans;
int PHchange;
int hash[20];
int biji;
struct Node
{
int effect;
int double_time;
int PHchange;
};
Node map[20];
void DFS(int now)
{
//printf("now is %d PHchange is %d sum is %d\n",now,map[now].PHchange,sum);
if(map[now].PHchange >= m)
{
ans = 1;
if(sum < minn)
minn = sum;
return ;
}
for(int i = 1;i <= n;i ++)
{
if(hash[i] == 0)
{
sum ++;
hash[i] = 1;
if(m - map[now].PHchange <= map[i].double_time)
map[i].PHchange = map[now].PHchange + map[i].effect * 2;
else
map[i].PHchange = map[now].PHchange + map[i].effect;
DFS(i);
sum --;
hash[i] = 0;
}
}
return ;
}
int main()
{
while(~scanf("%d%d",&n,&m))
{
for(int i = 0;i < 20;i ++)
{
map[i].effect = 0;
map[i].double_time = 0;
map[i].PHchange = 0;
}
memset(hash,0,sizeof(hash));
for(int i = 1;i <= n;i ++)
scanf("%d%d",&map[i].effect,&map[i].double_time);
minn = 0;flag = 0;ans = 0;
minn = 1000;
for(int i = 1;i <= n;i ++)
{
sum = 1;
biji = 0;
hash[i] = 1;
if(m <= map[i].double_time)
map[i].PHchange = map[i].effect * 2;
else
map[i].PHchange = map[i].effect;
DFS(i);
hash[i] = 0;
map[i].PHchange = 0;
}
if(ans == 0)
printf("-1\n");
else
printf("%d\n",minn);
}
}