标签:+= clu name getc size namespace style return sdi
题解见注释 (强行优化成 O(N))
/*
ans = (V[x] - E[xx]) - (V[y] - E[yy])
= (n mod 2) + E[yy] - E[xx]
= (n mod 2) + (E[yy] + 1/2 * E[xy]) - (E[xx] + 1/2 * E[xy])
= (n mod 2) + 1/2 sum_of_degree_in_y - 1/2 sum_of_degree_in_x
*/
#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int N=200005;
inline int read(){
int x=0; char ch=getchar();
for(;!isdigit(ch);ch=getchar());
for(;isdigit(ch);ch=getchar()) x=x*10+ch-‘0‘;
return x;
}
int n,d[N],ans,a[N];
int main(){
n=read();
for(int i=1;i<n;i++) d[read()]++,d[read()]++;
for(int i=1;i<=n;i++) a[d[i]]++;
for(int i=1,now=1;i<=n;i++) if(a[i]&1) ans+=(now&1)?-i:i,now^=1;
ans=(ans>>1)+(n&1),printf("%d\n",ans);
return 0;
}
标签:+= clu name getc size namespace style return sdi
原文地址:https://www.cnblogs.com/JYYHH/p/9275262.html