hdu 2094 产生冠军

像拓扑排序那样先搞好。然后只要判断一下这张图中入度为0的节点有几个，如果有1个就输出yes。0个或者1个以上的全部输出no。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<map>
#include<string>
#include<algorithm>
using namespace std;
int rudu[2005];
vector<int>abc[2005];
map<string, int>dd;
int main()
{
    int n, m, i, x, y, j, k, ji;
    while (~scanf("%d",  &m))
    {
        if (m == 0) break;
        n = 1;
        memset(rudu, 0, sizeof(rudu));
        for (i = 0; i < 2004; i++) abc[i].clear();
        dd.clear();
        char s1[1000], s2[1000];
        for (i = 0; i < m; i++)
        {
            scanf("%s%s", s1, s2);
            if (dd[s1] == 0){ dd[s1] = n; n++; }
            if (dd[s2] == 0){ dd[s2] = n; n++; }
            x = dd[s1];
            y = dd[s2];
            rudu[y]++;
            abc[x].push_back(y);
        }
        int sum = 0;
        for (i = 1; i <=1; i++)
        {
            sum = 0;
            for (j = 1; j <= n-1; j++)
            {
                if (rudu[j] == 0)
                {
                    sum++;
                    ji = j;
                }
            }
            if (sum > 1||sum==0) break;
            else
            {
                rudu[ji]--;
                for (k = 0; k < abc[ji].size(); k++) rudu[abc[ji][k]]--;
            }
        }
        if (i == 2) printf("Yes
");
        else printf("No
");
    }
    return 0;
}