D - Guess UVALive - 4255 拓扑排序

Given a sequence of integers, a1, a2, . . . , an, we define its sign matrix S such that, for 1 ≤ i ≤ j ≤ n, 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 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 (1 ≤ n ≤ 10), 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 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

拓扑排序：把sum(i,j)考虑成pre[j] - pre[i-1]的形式，然后可以建立图的关系，进行拓扑排序，从大到小，每一个拓扑序后面的点都比前面小，所以每当在拓扑排序里遇到该点（说明当前值比它大），都将遇到的点的sum 减一

注意i前缀序列是n+1项，【0，n】

#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
#include<string>
using namespace std;
#define MAXN 14
#define N 100
int in[MAXN], g[MAXN][MAXN];
int sum[MAXN];
char str[N];
int main()
{
    int T, n;
    scanf("%d", &T);
    while (T--)
    {
        memset(g, , sizeof(g));
        memset(in, , sizeof(in));
        memset(sum, , sizeof(sum));
        scanf("%d", &n);
        scanf("%s", str);
        int pos = ;
        for (int i = ; i <= n; i++)
        {
            for (int j = i; j <= n; j++)
            {
                char c;
                c = str[pos++];
                if (c == '+')
                {
                    g[j][i-] = ;
                    in[i - ]++;
                }
                else if(c == '-')
                {
                    g[i - ][j] = ;
                    in[j]++;
                }
            }
        }
        queue<int> q;
        while (!q.empty()) q.pop();
        for (int i = ; i <= n; i++)
            if (!in[i])
                q.push(i);
        while (!q.empty())
        {
            int tmp = q.front();
            q.pop();
            for (int i = ; i <= n; i++)//必须从0开始
            {
                if (g[tmp][i] == )
                {
                    sum[i]--;
                    if (--in[i] == )
                        q.push(i);
                }
            }
        }
        for (int i = ; i <= n; i++)
        {
            printf("%d", sum[i] - sum[i - ]);
            if (i != n)printf(" ");
            else printf("\n");
        }
    }
}