uva 1423 - Guess （topsort,4级）
Given a sequence of integers, a1, a2,..., an , we define its sign matrix S such that, for 1ijn , Sij = `` + " if ai +...+ aj > 0 ; Sij = `` - " if ai +...+ aj < 0 ; and Sij = ``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 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 -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':' ');
}
}
}