uva 1423 - Guess （topsort,4级）

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"

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

	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

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.

Write a program that, given a valid

Input

Your program is to read from standard input. The input consists of T test cases. The number of test cases Tis 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

Output

Your program is to write to standard output. For each test case, output exactly one line containing a sequence of n

Sample Input

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

Sample Output

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

思路：设f[i]=ans[0]+..+ans[i],那么上述的i行j列+,-,0关系就表示为f[j]>f[i-1],f[j]<f[i-1],f[j]==f[i-1];

来个拓扑排序，就好了，其中0比较讨厌，那就不管它，只要再拓扑排序时，将没有入度的点都置成一样的值，就一定满足了相等的情况。

 

#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
using namespace std;
const int mm=15;
vector<int> g[mm];
int id[mm],f[mm],ans[mm];
int cas,n;
void get(int n)
{ char c;
  memset(g,0,sizeof(g));
  memset(id,0,sizeof(id));
  for(int i=1;i<=n;++i)
    for(int j=i;j<=n;++j)
  {
    while(1)
    { c=getchar();
      if(c=='+'){g[j].push_back(i-1);++id[i-1];break;}
      else if(c=='-'){g[i-1].push_back(j);++id[j];break;}
      else if(c=='0'){/**g[j][i-1]=g[i-1][j]=2;*/break;}
    }
  }
}
void topsort()
{
  int top=-1,high=10;
  for(int i=0;i<=n;++i)
    if(!id[i])
    id[i]=top,top=i;
  for(int i=0;i<=n;++i)
  { int kop=-1;
    if(top==-1){return;}
    while(top!=-1)
    {
      int num=top;top=id[top];f[num]=high;
      int m=g[num].size();
      for(int i=0;i<m;++i)
      if(--id[g[num][i]]==0)
      {
       id[g[num][i]]=kop;kop=g[num][i];
      }
    }
    top=kop;
    high--;
  }
}
int main()
{
  while(~scanf("%d",&cas))
  {
    while(cas--)
    { scanf("%d",&n);
      get(n);
      topsort();
      for(int i=1;i<=n;++i)
        ans[i]=f[i]-f[i-1];
      for(int i=1;i<=n;++i)
        printf("%d%c",ans[i],i==n?'\n':' ');
    }
  }
}