UVA1220 - Party at Hali-Bula

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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 + 5map<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;}

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