bzoj 1131 简单树形dp

思路：随便想想就能想出来啦把。。。  卡了我一个vector。。。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int> >

using namespace std;

const int N = 1e6 + 10;
const int M = 10 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-6;

int n, tot, head[N];
LL cnt[N], sum[N], dp[N];

struct Edge {
    int to, nx;
} edge[N << 1];

void add(int u, int v) {
    edge[tot].to = v;
    edge[tot].nx = head[u];
    head[u] = tot++;
}

void dfs1(int u, int p) {
    cnt[u] = 1;
    for(int i = head[u]; ~i; i = edge[i].nx) {
        int v = edge[i].to;
        if(v == p) continue;
        dfs1(v, u);
        cnt[u] += cnt[v];
        sum[u] += sum[v] + cnt[v];
    }
}

void dfs2(int u, int p) {
    if(p) {
        dp[u] = sum[u] + (dp[p] - (sum[u] + cnt[u]) + n - cnt[u]);
    }

    for(int i = head[u]; ~i; i = edge[i].nx) {
        int v = edge[i].to;
        if(v == p) continue;
        dfs2(v, u);
    }
}

int main() {
    memset(head, -1, sizeof(head));
    scanf("%d", &n);
    for(int i = 1; i < n; i++) {
        int u, v; scanf("%d%d", &u, &v);
        add(u, v);
        add(v, u);
    }

    dfs1(1, 0);

    dp[1] = sum[1];
    dfs2(1, 0);

    int ans = 1;
    for(int i = 1; i <= n; i++) {
        if(dp[ans] < dp[i]) {
            ans = i;
        }
    }

    printf("%d\n", ans);
    return 0;
}
/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/9183525.html