BZOJ3251 树上三角形

一看这题。。。难道要链剖乱搞什么的吗。。。不会啊汗。。。

突然发现不构成三角形的条件其实非常苛刻，由斐波那契数列：

1,1,2,3,5,8,13,21,34......

可以知道其实小于int的大概就50项的样子。

于是路径长度>50直接输出'Y'，否则排序判断。。。

看来还是蛮快的。。。

/**************************************************************
    Problem: 3251
    User: rausen
    Language: C++
    Result: Accepted
    Time:268 ms
    Memory:5808 kb
****************************************************************/
 
#include <cstdio>
#include <algorithm>
 
using namespace std;
typedef long long ll;
const int N = 100005;
const int M = 200005;
struct edges{
    int next, to;
}e[M];
struct tree_node{
    int v, dep, fa;
}tr[N];
int first[N], q[N];
int n, m, tot;
 
inline int read(){
    int x = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9'){
        if (ch == '-') sgn = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9'){
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return sgn * x;
}
 
inline void add_edge(int x, int y){
    e[++tot].next = first[x];
    first[x] = tot;
    e[tot].to = y;
}
 
void add_Edges(int X, int Y){
    add_edge(X, Y);
    add_edge(Y, X);
}
 
void dfs(int p){
    int x, y;
    for (x = first[p]; x; x = e[x].next)
        if ((y = e[x].to) != tr[p].fa){
            tr[y].fa = p, tr[y].dep = tr[p].dep + 1;
            dfs(y);
        }
}
 
void query(int a, int b){
    int cnt = 0, i;
    while (cnt < 50 && a != b){
        if (tr[a].dep > tr[b].dep)
            q[++cnt] = tr[a].v, a = tr[a].fa;
        else q[++cnt] = tr[b].v, b = tr[b].fa;
    }
    q[++cnt] = tr[a].v;
    if (cnt >= 50){
        puts("Y");
        return;
    }
    sort(q + 1, q + cnt + 1);
    for(i = 1; i <= cnt - 2; ++i)
        if ((ll) q[i] + q[i + 1] > q[i + 2]){
            puts("Y");
            return;
        }
    puts("N");
}
 
int main(){
    n = read(), m = read();
    int i, T, X, Y;
    for (i = 1; i <= n; ++i)
        tr[i].v = read();
    for (i = 1; i < n; ++i){
        X = read(), Y = read();
        add_Edges(X, Y);
    }
    dfs(1);
    for (i = 1; i <= m; ++i){
        T = read(), X = read(), Y = read();
        if (T) tr[X].v = Y; else query(X, Y);
    }
    return 0;
}