UVA 1220Party at Hali-Bula

来源：互联网发布：淘宝虚拟物品发货编辑：程序博客网时间：2024/06/05 18:42

这个题还是很简单的、最主要的还是通过各种方法，把一个无根树变成有根树。然后进行搜索。

这个建树的过程很重要。用的是 map 标记和 vector 的下标。

然后再来实现就简单了。

#include <cstdio>#include <algorithm>#include <iostream>#include <cstring>#include <cmath>#include <cstdlib>#include <string>#include <map>#include <vector>#include <set>#include <queue>#include <stack>using namespace std;#define ll long long#define maxn 200+10#define INF 1<<30map <string , int> q;vector <string> fa;vector <int> son[maxn];int f[maxn][3] = {0};int init(){    q.clear();    fa.clear();    for(int i = 0; i < maxn; i++)        son[i].clear();    memset(f,0,sizeof(f));}int get_id(string  &s){    if(q.count(s))        return q[s];    fa.push_back(s);    return q[s] = fa.size() - 1;}int dp(int i,int j){    if(son[i].empty()){        f[i][j] = 1;        if(j == 0)            return 0;        return 1;    }    int v = 0;    if(j == 0){        int len = son[i].size();        int ok = 1;        for(int k = 0; k < len; k++){            int no_cho = dp(son[i][k] , 0),cho = dp(son[i][k] , 1);            v += max(no_cho , cho);            if(no_cho > cho ){                if(f[son[i][k]][0] != 1) ok = 0;            }            else if(no_cho < cho)                if(f[son[i][k]][1] != 1) ok=0;            else ok=0;        }        f[i][j]=(ok ? 1 : 0);    }    else{        v = 1;        int len = son[i].size(), ok = 1;        for(int k = 0; k < len; k++){            v += dp(son[i][k] , 0);            if(f[son[i][k]][0] != 1) ok=0;        }        f[i][j]=(ok ? 1 : 0);    }    return v;}int main (){    int n;    while(scanf("%d",&n) && n){        init();        string s;        cin >> s;        get_id(s); // 分配地址        for(int i = 1; i < n; i++){            string s1,s2;            cin >> s1 >> s2;            son[get_id(s2)].push_back(get_id(s1));        }        int no_cho = dp(0,0), cho = dp(0,1);        int flag = 1;//唯一        int max_;        if(no_cho == cho){            flag = 0;            max_ =cho;        }        else if(no_cho > cho){            if(f[0][0] != 1)                flag = 0;            max_ = no_cho;        }        else {            if(f[0][1] != 1)                flag = 0;            max_ = cho;        }        cout << max_ << (flag ? " Yes" : " No") << endl;    }    return 0;}

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