[bzoj1131][POI2008]Sta

1131: [POI2008]Sta

Time Limit: 10 Sec  Memory Limit: 162 MB
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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

sum表示子树，anti表示父亲作为子树的值

#include<bits/stdc++.h>using namespace std;const int N = 1000000 + 5;long long siz[N],anti[N],sum[N],ans; int ans2,last[N*2],cnt,n;struct Edge{ int to,next; }e[N*2];void insert( int u, int v ){e[++cnt].to = v; e[cnt].next = last[u]; last[u] = cnt;e[++cnt].to = u; e[cnt].next = last[v]; last[v] = cnt;}void dfs( int x, int f ){siz[x] = 1;for( int i = last[x]; i; i = e[i].next )if( e[i].to ^ f ){dfs( e[i].to, x );siz[x] += siz[e[i].to];sum[x] += (sum[e[i].to]+siz[e[i].to]);}}void dfs2( int x, int f ){for( int i = last[x]; i; i = e[i].next )if( e[i].to ^ f ){anti[e[i].to] = anti[x] + sum[x] - sum[e[i].to] - siz[e[i].to] + n - siz[e[i].to];dfs2( e[i].to, x );}if( anti[x] + sum[x] >= ans ){if( anti[x] + sum[x] > ans ) ans = anti[x] + sum[x], ans2 = x;else ans2 = min( ans2, x );}}int main(){scanf("%d", &n);for( int i = 1,u,v,w; i < n; i++ ){scanf("%d%d", &u, &v);insert( u, v );}dfs( 1, 0 ); dfs2( 1, 0 );printf("%d\n", ans2);return 0;}