D. Lizards and Basements 2 （DFS+最优性剪枝）

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http://codeforces.com/contest/6/problem/D

D. Lizards and Basements 2

This is simplified version of the problem used on the original contest. The original problem seems to have too difiicult solution. The constraints for input data have been reduced.

Polycarp likes to play computer role-playing game «Lizards and Basements». At the moment he is playing it as a magician. At one of the last levels he has to fight the line of archers. The only spell with which he can damage them is a fire ball. If Polycarp hits the i-th archer with his fire ball (they are numbered from left to right), the archer loses a health points. At the same time the spell damages the archers adjacent to the i-th (if any) — they lose b (1 ≤ b < a ≤ 10) health points each.

As the extreme archers (i.e. archers numbered 1 and n) are very far, the fire ball cannot reach them. Polycarp can hit any other archer with his fire ball.

The amount of health points for each archer is known. An archer will be killed when this amount is less than 0. What is the minimum amount of spells Polycarp can use to kill all the enemies?

Polycarp can throw his fire ball into an archer if the latter is already killed.

Input

The first line of the input contains three integers n, a, b (3 ≤ n ≤ 10; 1 ≤ b < a ≤ 10). The second line contains a sequence of n integers —h1, h2, ..., hn (1 ≤ hi ≤ 15), where hi is the amount of health points the i-th archer has.

Output

In the first line print t — the required minimum amount of fire balls.

In the second line print t numbers — indexes of the archers that Polycarp should hit to kill all the archers in t shots. All these numbers should be between 2 and n - 1. Separate numbers with spaces. If there are several solutions, output any of them. Print numbers in any order.

题意：你是火系法师，对面有一排敌人，每个敌人都有HP，你要向他们扔火球，火球对目标 i 有一个主伤害a，对i+1和i -1有一个溅射伤害b。法师只能攻击到2号到n-1号。问你至少要扔多少个火球才能杀死所有敌人，并依次输出每个火球的攻击目标。

题解：DFS暴搜+最优性剪枝。

从2号敌人到n-1号敌人进行搜索，攻击次数从至少杀死第 i - 1个敌人枚举到杀死i-1，i，i+1三个敌人。即h[x-1]<b*i。

DFS(x,times) ，x为目前攻击对象，times为总的攻击次数。

AC代码：

#include<bits/stdc++.h>
using namespace std;
int ans=9999999;
int h[100];
int a,b,n;
vector<int>V;
vector<int>V2;
void dfs(int x,int times)
{
	if(times>=ans)return;//最优性剪枝 
	if(x==n)//只能杀死 第 2 ~ n-1个人 
	{
		if(h[x]<0){
			V2=V;
			ans=times;
		}
		return ;
	}
	for(int i=0; i <= max( h[x-1]/b+1,max( h[x]/a+1, h[x+1]/b+1) );i++)//枚举次数 
	{
		if(h[x-1]<b*i)
		{
			h[x-1] -= b*i;
			h[x] -= a*i;
			h[x+1] -= b*i;
			for(int j=0;j<i;j++) V.push_back(x);
			dfs(x+1,times+i);
			for(int j=0;j<i;j++) V.pop_back();
			h[x-1] += b*i;
			h[x] += a*i;
			h[x+1] += b*i; 
		}
	}
}
int main()
{
	cin>>n>>a>>b;
	for(int i=1;i<=n;i++)cin>>h[i];
	dfs(2,0);
	cout<<ans<<endl;
	for(int i=0;i<V2.size();i++)cout<<V2[i]<<" ";
	cout<<endl;
	return 0;
}