cf1060E. Sergey and Subway(树形dp)

题意

题目链接

Sol

很套路的题

直接考虑每个边的贡献，最后再把奇数点的贡献算上

#include<bits/stdc++.h> 
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define LL long long 
#define rg register 
#define pt(x) printf("%d ", x);
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
#define chmin(x, y) (x = x < y ? x : y)
using namespace std;
const int MAXN = 1e6 + 10, INF = 1e18 + 10, mod = 1e9 + 7;
const double eps = 1e-9;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, dep[MAXN], num[MAXN], siz[MAXN], ans, a2;
vector<int> v[MAXN];
void dfs(int x, int fa) {
    dep[x] = dep[fa] + 1;
    num[dep[x] & 1]++;
    siz[x] = 1;
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i]; if(to == fa) continue;
        dfs(to, x);
        siz[x] += siz[to];
        
    }
    for(int i = 0; i < v[x].size(); i++) {
        int to = v[x][i]; if(to == fa) continue;
        ans += siz[to] * (N - siz[to]);	
    }
    
}
main() {
    N = read();
    for(int i = 1; i <= N - 1; i++) {
        int x = read(), y = read();
        v[x].push_back(y); v[y].push_back(x);
    }
    dfs(1, 0);
    for(int i = 1; i <= N; i++) a2 += num[(dep[i] & 1) ^ 1];
    printf("%lld", (ans + a2 / 2) / 2);
    return 0;	
}