HDU 2616 Kill the monster(DFS，类似全排列)

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在一芬菲的家乡附近有一座山，山上住着一只巨大的怪物。作为家乡的英雄，一芬菲决定消灭它。面对拥有大量生命值的怪物，一芬菲拥有多种咒语，每种咒语在不同的时机使用会有不同的效果。本文介绍了一芬菲如何通过最少的咒语使用次数来击败怪物的策略。

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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

此题不难，非常常规的dfs习题。每一次有不同的咒语选择，所以咒语的序号就构成了搜索树。并且每一次都可以选择n种咒语，感觉就是全排列衍伸而来的一道题，只不过全排列中是将1–n的数作为排列对象，这一题是将咒语作为排列对象！

代码：

#include<iostream>
using namespace std;

const int INF = 100;
const int maxn = 15;
int n, M;
int a[maxn];
int m[maxn];
int tag[maxn];
int min_res = INF;

void dfs(int k, int hp)		//hp表示怪兽此时的血量 
{
	if(hp <= 0)
	{
		if(k < min_res)
			min_res = k;
		return ;
	}
	for(int i = 1;i <= n;i++)		//每一次咒语的选择都是n种 
	{
		if(tag[i] != 1)		//如果咒语没有用过 
		{
			tag[i] = 1;
			if(hp <= m[i])
				dfs(k + 1, hp - a[i] * 2);
			else
				dfs(k + 1, hp - a[i]);
			tag[i] = 0;
		}
	}
}

void Clear()
{
	for(int i = 1;i <= n;i++)
	{
		tag[i] = 0;
	}
}

int main()
{
	while(scanf("%d%d", &n, &M) != EOF)
	{
		Clear();
		min_res = INF;
		for(int i = 1;i <= n;i++)
		{
			cin >> a[i] >> m[i];
		}
		dfs(0, M);
		if(min_res == INF)
			cout << "-1" << endl;
		else
			cout << min_res << endl;
	}
	return 0;
}