PerfectPan
;

hihocoder [Offer收割]编程练习赛98 占领树节点

April 04, 2019

题目链接http://hihocoder.com/contest/offers98/problem/4

题意:略。

思路:假如先手选了一个点,那么后手肯定是在它选的点周围选一个点作为起始点,因为如果不是相邻的话,先手就可以先把它们之间的路先堵上,这是不优的,然后问题就转化成了树上选一个点,这个点相邻的点的最大子树大小(后手可占据的最多点数)是否小于 nn- 最大子树的大小(先手可占据的最少点数),如果满足的话这个点就可选,时间复杂度 O(n)O(n)

#include <bits/stdc++.h>
#define MP make_pair
#define PB push_back
using namespace std;
typedef long long ll;
template<typename T>
inline T read(T&x){
    x=0;int f=0;char ch=getchar();
    while (ch<'0' || ch>'9') f|=(ch=='-'),ch=getchar();
    while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
    return x=f?-x:x;
}
const int N=1e5+10;
int n,i,j,fa[N],sz[N];
vector<int>G[N],vec;
void dfs(int u,int f){
    sz[u]=1;
    int mx=0;
    for (int i=0;i<(int)G[u].size();++i){
        int v=G[u][i];
        if (v==f) continue;
        dfs(v,u);
        sz[u]+=sz[v];
        mx=max(mx,sz[v]); 
    }
    mx=max(mx,n-sz[u]);
    if (mx<n-mx) vec.PB(u);
}
int main(){
    read(n);
    for (i=1;i<=n;++i){
        read(fa[i]);
        if(fa[i]){
            G[fa[i]].PB(i);
            G[i].PB(fa[i]);
        }
    }
    dfs(1,0);
    sort(vec.begin(),vec.end());
    printf("%d\n",(int)vec.size());
    for (i=0;i<(int)vec.size();++i){
        printf("%d\n",vec[i]);
    }
    return 0;
}

Yiming Pan

Written by Yiming Pan who lives and works in Hangzhou China. Welcome follow me on Github

Cannot load comments. Please check you network.