Nyoj 43 24 Point game 【DFS】

24 Point game

时间限制：3000 ms  |  内存限制：65535 KB

难度：5

描写叙述

There is a game which is called 24 Point game.

In this game , you will be given some numbers. Your task is to find an expression which have all the given numbers and the value of the expression should be 24 .The expression mustn't have any other operator except plus,minus,multiply,divide and the brackets. 

e.g. If the numbers you are given is "3 3 8 8", you can give "8/(3-8/3)" as an answer. All the numbers should be used and the bracktes can be nested. 

Your task in this problem is only to judge whether the given numbers can be used to find a expression whose value is the given number。

输入

The input has multicases and each case contains one line
The first line of the input is an non-negative integer C(C<=100),which indicates the number of the cases.
Each line has some integers,the first integer M(0<=M<=5) is the total number of the given numbers to consist the expression,the second integers N(0<=N<=100) is the number which the value of the expression should be.
Then,the followed M integer is the given numbers. All the given numbers is non-negative and less than 100

输出

For each test-cases,output "Yes" if there is an expression which fit all the demands,otherwise output "No" instead.

例子输入

2
4 24 3 3 8 8
3 24 8 3 3

例子输出

Yes
No

分析：能够进行的运算为加。 减， 被减， 乘， 除。 被除六种运算。我们仅仅须要每次都取出没实用的两个数（能够是运算之后的值），进行运算，枚举就好了。

代码：

 
#include <cstdio>
#include <cstring>
#include <cmath>
const double E = 1e-6;
using namespace std;

double s[50], res;  
bool vis[50];

int dfs(int m, int top){  //每次取出两个数然后将结果放入数组中，就等于每次减掉一个没有使用的数
	if(m == 1){
		if(fabs(res - s[top-1]) < E){   
			//printf("%lf...res %lf..sll
", s[top-1], res);
			return 1;
			
		}
		return 0;
	}
	for(int i = 0; i < top-1; ++ i){
		if(!vis[i]){
			vis[i] = 1;
			for(int j = i+1; j < top; ++ j){
				if(!vis[j]){
					vis[j] = 1;
					
					s[top] = s[i]+s[j];
					if(dfs(m-1, top+1)) return 1;
					
					s[top] = s[i]-s[j];
					if(dfs(m-1, top+1)) return 1;
					
					s[top] = s[j]-s[i];
					if(dfs(m-1, top+1)) return 1;
					
					s[top] = s[i]*s[j];
					if(dfs(m-1, top+1)) return 1;
					
					if(s[i] != 0){
						s[top] = s[j]/s[i];
						if(dfs(m-1, top+1)) return 1;
					}
					
					if(s[j] != 0){
						s[top] = s[i]/s[j];
						if(dfs(m-1, top+1)) return 1;
					}
					
					vis[j] = 0;
				}
			}
			vis[i] = 0;
		}
	}
	return 0;
}

int main(){
	int t;
	scanf("%d", &t);
	while(t --){
		int n;
		scanf("%d", &n);
		scanf("%lf", &res);
		for(int i = 0; i < n; ++ i) scanf("%lf", &s[i]);
		memset(vis, 0, sizeof(vis));
		if(dfs(n, n)) printf("Yes
");
		else printf("No
");
	}
	return 0;
}