[DFS&&剪枝]uva10400 Game Show Math

Problem H
Game Show Math
Input: standard input
Output: standard output
Time Limit: 15 seconds

A game show in Britain has a segment where it gives its contestants a sequence of positive numbers and a target number. The contestant must make a mathematical expression using all of the numbers in the sequence and only the operators: +, -, *, and, /. Each number in the sequence must be used exactly once, but each operator may be used zero to many times. The expression should be read from left to right, without regard for order of operations, to calculate the target number. It is possible that no expression can generate the target number. It is possible that many expressions can generate the target number.

There are three restrictions on the composition of the mathematical expression:

o  the numbers in the expression must appear in the same order as they appear in the input file

o  since the target will always be an integer value (a positive number), you are only allowed to use / in the expression when the result will give a remainder of zero.

o  you are only allowed to use an operator in the expression, if its result after applying that operator is an integer from (-32000 ..+32000).

Input

The input file describes multiple test cases. The first line contains the number of test cases n.

Each subsequent line contains the number of positive numbers in the sequence p, followed by p positive numbers, followed by the target number. Note that 0 < p £ 100. There may be duplicate numbers in the sequence. But all the numbers are less than 32000.

Output

The output file should contain an expression, including all k numbers and (k-1) operators plus the equals sign and the target. Do not include spaces in your expression. Remember that order of operations does not apply here. If there is no expression possible output "NO EXPRESSION" (without the quotes). If more than one expression is possible, any one of them will do.

 

Sample Input

3
3 5 7 4 3
2 1 1 2000
5 12 2 5 1 2 4

Sample Output
5+7/4=3 
NO EXPRESSION
12-2/5*1*2=4

---

(Problem-setter: Sandy Graham, CS Dept, University of Waterloo)

 

题意：

在英国有一个数学游戏，给参赛者一些正整数和一个目标数，参赛者必须在这些正整数间插入+、-、*或/ 的符号，使得最后计算的结果等于目标数。计算的方式是由左到右，而且不必管运算的优先顺序（就是不管先乘除后加减那一套）。

在这个数学运算式中，有三个限制：

• 正整数出现的次序不可改变，也就是要与输入的顺序相同
• 因为目标数也是一个正整数，所以在运算的过程中，你只有在可以整除的情况下才可以使用/ 。
• 在运算的过程中，如果你用某一​​个运算符号，会导致产生的数超出（-32000 ～ +32000）的范围，那么你不可以采用此运算符号。（也就是说在运算的过程中都不该有超出范围的数出现）

思路：典型的DFS了，但是由于数据量较大，必须要剪枝，把所有的状态记录在vis数组里面，记为vis[100][64000]，意思是在某一步表达式的值，遇到已经在同一步求出来的就可以认为不必往下面搜索了。。想了半天没想出来。。

#include<iostream>
#include<cstring>

using namespace std;

int arry[110],vis[110][64100];
int tag,res,m;
char vec[150];

void dfs(int pos,int cnt)
{
    if(tag) return;
    if(pos+1==m&&cnt==res)
    {
        tag=1;
        return;
    }
    else if(pos+1==m) return;
    if(cnt+arry[pos+1]<=32000&&cnt+arry[pos+1]>=-32000&&tag==0&&vis[pos+1][cnt+arry[pos+1]+32000]==0)
    {
		vis[pos+1][cnt+arry[pos+1]+32000]=1;
        dfs(pos+1,cnt+arry[pos+1]);
        if(tag)
        {
			vec[pos+1]='+';
            return;
        }
    }
    if(cnt-arry[pos+1]<=32000&&cnt-arry[pos+1]>=-32000&&tag==0&&vis[pos+1][cnt-arry[pos+1]+32000]==0)
    {
		vis[pos+1][cnt-arry[pos+1]+32000]=1;
        dfs(pos+1,cnt-arry[pos+1]);
        if(tag)
        {
			vec[pos+1]='-';
			return;
        }
    }

    if(cnt%arry[pos+1]==0&&cnt/arry[pos+1]<=32000&&cnt/arry[pos+1]>=-32000&&tag==0&&vis[pos+1][cnt/arry[pos+1]+32000]==0)
    {
		vis[pos+1][cnt/arry[pos+1]+32000]=1;
        dfs(pos+1,cnt/arry[pos+1]);
        if(tag)
        {
			vec[pos+1]='/';
			return;
        }
    }
	if(cnt*arry[pos+1]<=32000&&cnt*arry[pos+1]>=-32000&&tag==0&&vis[pos+1][cnt*arry[pos+1]+32000]==0)
    {
		vis[pos+1][cnt*arry[pos+1]+32000]=1;
        dfs(pos+1,cnt*arry[pos+1]);
        if(tag)
        {
			vec[pos+1]='*';
			return;
        }
    }
    return;
}

int main()
{
    int n;
    cin>>n;
    while(n--)
    {
        int i;
        tag=0;
        cin>>m;
        for(i=0;i<m;i++)
            cin>>arry[i];
        cin>>res;
		memset(vis,0,sizeof(vis));
        dfs(0,arry[0]);
        if(tag)
        {
            for(i=0;i<m-1;i++)
            {
                cout<<arry[i]<<vec[i+1];
            }
            cout<<arry[i]<<"="<<res<<endl;
        }
        else cout<<"NO EXPRESSION"<<endl;
    }
    return 0;
}