Codeforces Gym 100650B Countdown DFS

题目链接:

http://codeforces.com/gym/100650/attachments

题意:

给你n个父子关系，然后让你求出前三多的第d代儿子的人是谁

题解:

建一棵树。dfs搞一搞

代码:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define MS(a) memset(a,0,sizeof(a))
#define MP make_pair
#define PB push_back
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
inline ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
//////////////////////////////////////////////////////////////////////////
const int maxn = 1e3+10;

string s1,s2;
vector<int> E[maxn];
map<string,int> mp;
int tot;

struct node{
    string name;
    int num;
    bool operator<(const node &rhs)const{
        if(num == rhs.num)
            return name < rhs.name;
        return num > rhs.num;
    }
}fam[maxn];

int getID(string s){
    if(mp[s]==0){
        mp[s] = ++tot;
        fam[tot].name = s;
        fam[tot].num = 0;
    }
    return mp[s];
}

int solve(int x,int d){
    if(d == 0)
        return 1;

    int ans = 0;
    for(int i=0; i<(int)E[x].size(); i++){
        ans += solve(E[x][i],d-1);
    }
    return ans;
}

int main(){
    int T=read();
    for(int cas=1; cas<=T; cas++){
        if(cas!=1) cout << endl;
        mp.clear();
        for(int i=0; i<=maxn; i++) E[i].clear();
        tot = 0;
        int n=read(),d=read();
        for(int i=0; i<n; i++){
            cin >> s1;
            int k = read();
            for(int j=0; j<k; j++){
                cin >> s2;
                E[getID(s1)].push_back(getID(s2));
            }
        }

        for(int i=1; i<=tot; i++){
            fam[i].num = solve(i,d);
        }
        sort(fam+1,fam+1+tot);

        int i,tmp=0;
        // for(int i=1; i<=tot; i++){
        //  cout << fam[i].name << " " << fam[i].num << endl;
        // }
        cout << "Tree " << cas << ":
";
        for(i=1; i<=min(3,tot); i++){
            if(fam[i].num == 0) break;
            cout << fam[i].name << " " << fam[i].num << endl;
            tmp = fam[i].num;
        }
        // cout << "oo " << tmp << endl;
        for(int j=i; j<=tot; j++){
            if(fam[j].num<tmp || fam[j].num==0) break;
            cout << fam[j].name << " " << fam[j].num << endl;
        }
    }

    return 0;
}