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。。。

 1 /**************************************************************
 2     Problem: 3251
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:268 ms
 7     Memory:5808 kb
 8 ****************************************************************/
 9
10 #include <cstdio>
11 #include <algorithm>
12
13 using namespace std;
14 typedef long long ll;
15 const int N = 100005;
16 const int M = 200005;
17 struct edges{
18     int next, to;
19 }e[M];
20 struct tree_node{
21     int v, dep, fa;
22 }tr[N];
23 int first[N], q[N];
24 int n, m, tot;
25
26 inline int read(){
27     int x = 0, sgn = 1;
28     char ch = getchar();
29     while (ch < ‘0‘ || ch > ‘9‘){
30         if (ch == ‘-‘) sgn = -1;
31         ch = getchar();
32     }
33     while (ch >= ‘0‘ && ch <= ‘9‘){
34         x = x * 10 + ch - ‘0‘;
35         ch = getchar();
36     }
37     return sgn * x;
38 }
39
40 inline void add_edge(int x, int y){
41     e[++tot].next = first[x];
42     first[x] = tot;
43     e[tot].to = y;
44 }
45
46 void add_Edges(int X, int Y){
47     add_edge(X, Y);
48     add_edge(Y, X);
49 }
50
51 void dfs(int p){
52     int x, y;
53     for (x = first[p]; x; x = e[x].next)
54         if ((y = e[x].to) != tr[p].fa){
55             tr[y].fa = p, tr[y].dep = tr[p].dep + 1;
56             dfs(y);
57         }
58 }
59
60 void query(int a, int b){
61     int cnt = 0, i;
62     while (cnt < 50 && a != b){
63         if (tr[a].dep > tr[b].dep)
64             q[++cnt] = tr[a].v, a = tr[a].fa;
65         else q[++cnt] = tr[b].v, b = tr[b].fa;
66     }
67     q[++cnt] = tr[a].v;
68     if (cnt >= 50){
69         puts("Y");
70         return;
71     }
72     sort(q + 1, q + cnt + 1);
73     for(i = 1; i <= cnt - 2; ++i)
74         if ((ll) q[i] + q[i + 1] > q[i + 2]){
75             puts("Y");
76             return;
77         }
78     puts("N");
79 }
80
81 int main(){
82     n = read(), m = read();
83     int i, T, X, Y;
84     for (i = 1; i <= n; ++i)
85         tr[i].v = read();
86     for (i = 1; i < n; ++i){
87         X = read(), Y = read();
88         add_Edges(X, Y);
89     }
90     dfs(1);
91     for (i = 1; i <= m; ++i){
92         T = read(), X = read(), Y = read();
93         if (T) tr[X].v = Y; else query(X, Y);
94     }
95     return 0;
96 }