Bzoj-1131 Sta [DFS]

1131: [POI2008]Sta

Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 1629  Solved: 611
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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

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比较简单的一题当我们在u点时   假设u为根的树深度和为X  那么向u的子节点v走一步后  u及u的祖先和v的兄弟等所有不属于v所在子树的节点的深度都+1   而v所在子树的所有点包括v的深度都-1  所以X变成了  X = X + (n - siz[v]) - siz[v]所以两次DFS就可以出答案了防爆栈写了BFS  但其实旁边的大佬用DFS也A了

/**************************************************************    Problem: 1131    User:     Language: C++    Result: Accepted    Time:6760 ms    Memory:44264 kb****************************************************************/ #include <bits/stdc++.h>using namespace std; const int N = 1e6 + 137; inline void read( int &res ){    char ch = getchar(); res = 0;    while(ch < '0' || ch > '9') ch = getchar();    while(ch >= '0' && ch <= '9') res = res * 10 + ch - 48, ch = getchar();} int n, head[N], top; struct Node{    int y, nxt;    Node() {    }    Node( int y, int nxt ) : y(y), nxt(nxt) {   }} e[N << 1]; inline void Adde( int x, int y ){    e[++top] = Node(y, head[x]), head[x] = top;    e[++top] = Node(x, head[y]), head[y] = top;} int que[N], h, t, fa[N], siz[N], dep[N]; void Bfs(){    que[++t] = 1;    fa[1] = 0;    dep[1] = 0;    while(h < t)    {        int u = que[++h];        siz[u] = 1;        for(int i = head[u]; i; i = e[i].nxt)        {            int v = e[i].y;            if(v == fa[u]) continue;            fa[v] = u;            dep[v] = dep[u] + 1;            que[++t] = v;        }    }    for(int i = t; i >= 1; --i)        for(int j = head[que[i]]; j; j = e[j].nxt)            if(e[j].y != fa[que[i]]) siz[que[i]] += siz[e[j].y];} long long f[N], ans = 0, pos; void Bfs1(){    h = t = 0;    que[++t] = 1;    for(int i = 1; i <= n; ++i) f[1] += dep[i];    ans = f[1];    pos = 1;    while(h < t)    {        int u = que[++h];        for(int i = head[u]; i; i = e[i].nxt)        {            int v = e[i].y;            if(v == fa[u]) continue;            f[v] = f[u] + (n - siz[v]) - siz[v];            if(f[v] > ans || (f[v] == ans && pos > v))                ans = f[v], pos = v;            que[++t] = v;        }    }} int main(){    read(n);    for(int i = 1, u, v; i < n; ++i)        read(u), read(v), Adde(u, v);    Bfs();    Bfs1();    cout << pos << endl;    return 0;}/*81 45 64 56 76 82 43 4*/