拓扑排序简单题

产生冠军

题目传送：HDU - 2094 - 产生冠军

思路：因为只要确定是否产生冠军，根据题意可以得知当且仅当入度为0的只有一个时可以产生冠军，可以用map来映射存储的人的名字

AC代码：

#include <map>
#include <set>
#include <cmath>
#include <deque>
#include <queue>
#include <stack>
#include <cstdio>
#include <cctype>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define INF 0x7fffffff
using namespace std;

map<string, int> mp;
int deg[1005];
int cnt;
int n;

int main() {
    while(scanf("%d", &n) != EOF) {
        if(n == 0) break;
        memset(deg, 0, sizeof(deg));
        cnt = 1;
        mp.clear();

        string s1, s2;
        for(int i = 0; i < n; i ++) {
            cin >> s1 >> s2;
//          cout << s1 << " " << s2 << endl;
            if(mp.find(s1) == mp.end()) {
                mp[s1] = cnt ++;
            }
            if(mp.find(s2) == mp.end()) {
                mp[s2] = cnt ++;
            }
            int t1 = mp[s1];
            int t2 = mp[s2];
            deg[mp[s2]] ++;
        }

        int sum = 0;
        for(int i = 1; i < cnt; i ++) {
            if(deg[i] == 0) {
                sum ++;
            }
        }

        if(sum == 1) {
            printf("Yes\n");
        }
        else printf("No\n");
    }
    return 0;
}