51Nod 1405 树的距离之和 (树dp)

题目链接：http://www.51nod.com/onlineJudge/questionCode.html#!problemId=1405

中文题面不解释了，两次dfs，第一次自下向上，第二次自上向下。

ans[i]表示i节点的答案，cnt[i]表示i节点为root的子树的节点个数，d[i]表示i节点为root的子树的答案。

//#pragma comment(linker, "/STACK:102400000, 102400000")
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <list>
#include <set>
#include <map>
using namespace std;
typedef long long LL;
typedef pair <int, int> P;
const int N = 1e5 + 5;
LL cnt[N], d[N], ans[N], n;
vector <int> edge[N];

void dfs1(int u, int p) {
    cnt[u] = 1;
    d[u] = 0;
    for(int i = 0; i < edge[u].size(); ++i) {
        int v = edge[u][i];
        if(v == p)
            continue;
        dfs1(v, u);
        cnt[u] += cnt[v];
        d[u] += cnt[v] + d[v];
    }
}

void dfs2(int u, int p) {
    if(p != -1) {
        ans[u] = (ans[p] - d[u] - cnt[u]) + (n - cnt[u]) + d[u];
    } else {
        ans[u] = d[u];
    }
    for(int i = 0; i < edge[u].size(); ++i) {
        int v = edge[u][i];
        if(v == p)
            continue;
        dfs2(v, u);
    }
}

int main()
{
    int u, v;
    while(~scanf("%lld", &n)) {
        for(int i = 1; i <= n; ++i) {
            edge[i].clear();
        }
        for(int i = 1; i < n; ++i) {
            scanf("%d %d", &u, &v);
            edge[u].push_back(v);
            edge[v].push_back(u);
        }
        dfs1(1, -1);
        dfs2(1, -1);
        for(int i = 1; i <= n; ++i) {
            printf("%lld
", ans[i]);
        }
    }
    return 0;
}