BZOJ 1131: [POI2008]Sta( dfs )

 对于一棵树, 考虑root的答案向它的孩子转移, 应该是 ans[son] = (ans[root] - size[son]) + (n - size[son]).

so , 先 dfs 预处理一下, 然后再 DFS 求出各点答案 , 取最优即可 

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#include<cstdio>

#include<cstring>

#include<algorithm>

#include<iostream>

 

#define rep(i ,n) for(int i=0; i < n; ++i)

#define clr(x ,c) memset(x, c, sizeof(x))

 

using namespace std;

 

typedef long long ll;

 

const int maxn = 1000005;

 

struct edge {

int to;

edge*next;

} E[maxn << 1], *pt = E, *head[maxn];

 

void add(int u, int v) {

pt->to = v;

pt->next = head[u];

head[u] = pt++;

}

#define add_edge(u, v) add(u, v), add(v, u)

 

int size[maxn], dep[maxn], n;

ll ans[maxn];

 

void init() {

clr(head, 0);

cin >> n;

rep(i, n - 1) {

int u, v;

scanf("%d%d", &u, &v);

u--, v--;

add_edge(u, v);

}

}

 

void dfs(int x, int fa) {

size[x] = 1;

for(edge*e= head[x]; e; e = e->next) if(e->to != fa) {

ans[0] += (dep[e->to] = dep[x] + 1);

dfs(e->to, x);

size[x] += size[e->to];

}

}

 

void DFS(int x, int fa) {

for(edge*e = head[x]; e; e = e->next) if(e->to != fa) {

ans[e->to] = ans[x] + n - 2 * size[e->to];

DFS(e->to, x);

}

}

 

void work() {

ans[0] = dep[0] = 0;

dfs(0, -1);

DFS(0, -1);

cout << max_element(ans, ans + n) - ans + 1 << "\n";

}

 

int main(){

freopen( "test.in" , "r" , stdin );

init();

work();

return 0;

} 

 

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1131: [POI2008]Sta

Time Limit: 10 Sec   Memory Limit: 162 MB
Submit: 870   Solved: 271
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Description

给出一个N个点的树,找出一个点来,以这个点为根的树时,所有点的深度之和最大

Input

给出一个数字N,代表有N个点.N<=1000000 下面N-1条边.

Output

输出你所找到的点,如果具有多个解,请输出编号最小的那个.

Sample Input

8
1 4
5 6
4 5
6 7
6 8
2 4
3 4

Sample Output

7

HINT

Source