UVA1220 - Party at Hali-Bula

这道题几乎就是树的最大独立集问题，只不过要判断唯一性。
在这里用了个ju[0], ju[1],ju[2]，来判断相应的d, s, gs数组中值得唯一性.

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <map>
#include <vector>
#include <cmath>
using namespace std;

#define maxn 200 + 5
map<string, int> m;
vector<int> v[maxn];
int d[maxn], s[maxn], gs[maxn], ju[3][maxn];
void dp(int n, int fa)
{
    if(v[n].empty())
    {
        d[n] = 1;
        return ;
    }
    for(int i = 0; i < v[n].size(); i++)
    {
        int h = v[n][i];
        dp(h, n);
        s[n] += d[h];
        if(ju[0][h])
            ju[1][n] = true;
        if(fa != -1)
        {
            gs[fa] += d[h];
            if(ju[0][h])
                ju[2][fa] = true;
        }
    }
    if(s[n] == gs[n] + 1)
    {
        d[n] = s[n];
        ju[0][n] = true;
    }
    else if(s[n] > gs[n] + 1)
    {
        d[n] = s[n];
        if(ju[1][n])
            ju[0][n] = true;
    }
    else {
        d[n] = gs[n] + 1;
        if(ju[2][n])
            ju[0][n] = true;
    }
}
int main()
{
   // freopen("in.txt", "r", stdin);
    int n;

    while(cin >> n && n)
    {
        string s1, s2;
        int cnt = 0;

        m.clear();
        for(int i = 0; i <= maxn; i++)
            v[i].clear();
        cin >> s1;
        m[s1] = cnt++;
        for(int i = 1; i < n; i++)
        {
            cin >> s1 >> s2;
            if(!m.count(s1))
                m[s1] = cnt++;
            if(!m.count(s2))
                m[s2] = cnt++;
            v[m[s2]].push_back(m[s1]);
        }
        memset(s, 0, sizeof(s));
        memset(gs, 0, sizeof(gs));
        memset(ju, 0, sizeof(ju));
        dp(0, -1);
        cout << d[0] << " ";
        if(ju[0][0])
            cout << "No" << endl;
        else
            cout << "Yes" << endl;
    }

    return 0;
}