UVALive - 4255 Guess

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题意：对于一个序列，我们可以计算出一个符号矩阵，其中Sij为ai+...+aj的正负号，

现在给你一个矩阵的上三角，求一个满足的序列

思路：如果Sij>0的话，那么代表前缀和差Bj-Bi-1 >0 ,那么Bj > Bi-1，由此我们可以得到一系列的关系，利用toposort排序后，得到一个递增或者递减的序列，就可以求出来各个数了

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>using namespace std;const int MAXN = 20;int n,degree[MAXN];int map[MAXN][MAXN],s[MAXN],ans[MAXN];char text[1000];void init(){    memset(map,0,sizeof(map));    memset(degree,0,sizeof(degree));}void toposort(){    int count = 0,low = -10;    while (count != n+1){        int tag[MAXN];        memset(tag,0,sizeof(tag));        for (int i = 0; i <= n; i++)            if (degree[i] == 0){                s[i] = low;                degree[i] = -1;                tag[i] = 1;                count++;            }        low++;        for (int i = 0; i <= n; i++)            if (tag[i]){                for (int j = 0; j <= n; j++)                    if (map[i][j])                        degree[j]--;            }    }}int main(){    int t;    scanf("%d",&t);    while (t--){        scanf("%d%s",&n,text);        init();        int cnt = 0;        for (int i = 1; i <= n; i++)            for (int j = i; j <= n; j++){                char a = text[cnt++];                if (a == '+'){                    map[i-1][j] = 1;                    degree[j]++;                }                else if (a == '-'){                    map[j][i-1] = 1;                    degree[i-1]++;                }            }        toposort();        for (int i = 1; i <= n; i++)            ans[i] = s[i] - s[i-1];        for (int i = 1; i < n; i++)            printf("%d ",ans[i]);        printf("%d\n",ans[n]);    }    return 0;}

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