[拓扑排序] PKU 1094 Sorting It All Out

第一道拓扑排序，拓扑排序的过程很简单：每次拿出一个入度为0的顶点放入solu[]尾部，并将它可以直接到达的顶点的入度减1；

对这道题需要得到唯一顺序，每次发现有多个入度为0的顶点时，说明还未排好序，此时不应该直接返回cannot be determined而是对所有入度为0的顶点执行上述操作，如果发现到某一步没有入度为0的顶点了说明出现矛盾，此时返回出现矛盾；

PS:这道题的输入有重复。

# include <cstdio>
# include <cstring>

# define N 26 + 1

int n, m;
int in[N], tmp[N];
char ans[N], g[N][N];

int topsort(void)
{
    bool multi = false;
    int cnt, i, first;
     for (i = 0; i < n; ++i)
         tmp[i] = in[i];
     char vis[N];
     memset(vis, 0, sizeof(char)*n);
     int c = 0;
    while(c < n)
    {
        cnt = 0;
        for (i = 0; i < n; ++i)    if (!vis[i] && tmp[i] == 0)
        {
            ++cnt;
            if (cnt > 1)
            {
                multi = true;
                break;
            }
            else
            {
                vis[i] = 1;
                first = i;
            }
        }
        if (cnt == 0) return -1;
        for (i = 0; i < n; ++i) if (g[first][i])
            --tmp[i];
        if (multi == false)
            ans[c] = first+'A';
        ++c;
     }
     if (multi == true) return 1;
     ans[c] = 0;
    return 0;
}

void solve(void)
{
    char s[5];
    int t, u, v;
    bool finished = false;
    for (int i = 1; i <= m; ++i)
    {
        scanf("%s", s);
        u = s[0]-'A', v = s[2]-'A';
        if (!g[u][v]) g[u][v] = 1, ++in[v];
        if (finished == false)
        {
            t = topsort();
            if (t == 0)
            {
                printf("Sorted sequence determined after %d relations: %s.\n", i, ans);
                finished = true ;
            }
            else if (t == -1)
            {
                printf("Inconsistency found after %d relations.\n", i);
                finished = true ;
            }
        }
    }
    if (finished == false) printf("Sorted sequence cannot be determined.\n");
}

void init(void)
{
    for (int i = 0; i < n; ++i)
    {
        in[i] = 0;
        memset(g[i], 0, sizeof(char)*n);
    }
}

int main()
{    
    while (scanf("%d%d", &n, &m), n||m)
    {
        init();
        solve();
    }
    
    return 0;
}