TAT~图论好难，3个月前就见过这道题了，当初拿过来这题根本没感觉是个图论。

而后听了嘉神简单讲了下思路也是糊里糊涂，只知道是个拓扑排序问题。如今再看到这题依旧没个敲的思路。

直到盯着队友的代码看了半个小时，又手跑了一趟，才终于明白个所以然，感觉好神奇= =|||

【虽然能感觉到自己有进步，但真不知道什么时候才能把自己这颗“雪球”滚的越来越大】

Description

Given a sequence of integers, a₁, a₂,..., a_n , we define its sign matrix S such that, for 1ijn , S_ij = `` + " if a<sub>i</sub> +...+ a<sub>j</sub> > 0 ; S<sub>ij</sub> = `` - "if a_i +...+ a_j < 0 ; and S_ij = ``0" otherwise.

For example, if (a₁, a₂, a₃, a₄) = (- 1, 5, - 4, 2) , then its sign matrix S is a 4×4 matrix:

	1	2	3	4
1	-	+	0	+
2		+	+	+
3			-	-
4				+

We say that the sequence (-1, 5, -4, 2) generates the sign matrix. A sign matrix is valid if it can be generated by a sequence of integers.

Given a sequence of integers, it is easy to compute its sign matrix. This problem is about the opposite direction: Given a valid sign matrix, find a sequence of integers that generates the sign matrix. Note that two or more different sequences of integers can generate the same sign matrix. For example, the sequence (-2, 5, -3, 1) generates the same sign matrix as the sequence (-1,5, -4,2).

Write a program that, given a valid sign matrix, can find a sequence of integers that generates the sign matrix. You may assume that every integer in a sequence is between -10 and 10, both inclusive.

Input

Your program is to read from standard input. The input consists of T test cases. The number of test cases T is given in the first line of the input. Each test case consists of two lines. The first line contains an integer n(1n10) , where n is the length of a sequence of integers. The second line contains a string of n(n + 1)/2 characters such that the first n characters correspond to the first row of the sign matrix, the next n - 1 characters to the second row, ... , and the last character to the n -th row.

Output

Your program is to write to standard output. For each test case, output exactly one line containing a sequence of n integers which generates the sign matrix. If more than one sequence generates the sign matrix, you may output any one of them. Every integer in the sequence must be between -10 and 10, both inclusive.

Sample Input

3 4 -+0++++--+ 2 +++ 5 ++0+-+-+--+-+--

Sample Output

-2 5 -3 1 3 4 1 2 -3 4 -5

解题思路：首先是要把求每个位置上的值转化为求 “前缀和之差”。

input数组中，input[i]中存的值，表示有多少个"前缀和"比"前i个数的和"大。

vec数组中，vec[i]中存的值x，表示"前x数的和"比"前i个小"。

然后假设最大的前缀和为20，依次第二大19,第三大18递减得到。

#include 
    
     #include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           using namespace std;#define maxn 15char ch[maxn * maxn];int t, n, k, ans;int input[maxn], sum[maxn];vector
           
             vec[maxn];void init() { k = 0; ans = 20; memset(input, 0, sizeof(input)); memset(sum, 0, sizeof(sum)); for(int i = 0; i <= n; i++) vec[i].clear();}void add_edge() { for(int i = 1; i <= n; i++) { for(int j = i; j <= n; j++) { char temp = ch[k++]; if(temp == '+') { input[i - 1]++; vec[j].push_back(i - 1); } else if(temp == '-') { input[j]++; vec[i - 1].push_back(j); } } }}void topsort() { stack
            
              s; for(int i = 0; i <= n; i++) if(input[i] == 0) { s.push(i); sum[i] = ans; input[i] = -1; } while(!s.empty()) { int x = s.top(); s.pop(); ans--; input[x] = -1; for(int i = 0; i < vec[x].size(); i++) { input[vec[x][i]]--; if(input[vec[x][i]] == 0) { s.push(vec[x][i]); input[vec[x][i]] = -1; sum[vec[x][i]] = ans; } } }}int main() { scanf("%d", &t); while(t--) { scanf("%d%s", &n, ch); init(); add_edge(); topsort(); for(int i = 1; i <= n; i++) { if(i == 1) printf("%d", sum[i] - sum[i - 1]); else printf(" %d", sum[i] - sum[i - 1]); } printf("\n"); } return 0;}