洛谷p3398仓鼠找suger题解

我现在爱死树链剖分了

题目

具体分析的话在洛谷blog

这里只是想放一下改完之后的代码

多了一个son数组少了一个for

少了找size最大的儿子的for

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = ;
int n, q, head[N], cnt, dad[N], top[N], size[N], deep[N], son[N];
struct edge{
int next, to;
}e[N << ];
int read() {
int s = , w = ;
char ch = getchar();
while(!isdigit(ch)) {if(ch == '-') w = -;ch = getchar();}
while(isdigit(ch)) {s = s * + ch - '';ch = getchar();}
return s * w;
}
void add(int x, int y) {
e[++cnt].next = head[x];
e[cnt].to = y;
head[x] = cnt;
}
void dfs(int now) {
size[now] = ;
deep[now] = deep[dad[now]] + ;
for(int i = head[now]; i; i = e[i].next)
if(e[i].to != dad[now]) {
dad[e[i].to] = now, dfs(e[i].to), size[now] += size[e[i].to];
if(size[e[i].to] > size[son[now]]) son[now] = e[i].to;
}
}
void dfs1(int now) {
if(!top[now]) top[now] = now;
if(son[now]) top[son[now]] = top[now], dfs1(son[now]);
for(int i = head[now]; i; i = e[i].next)
if(e[i].to != dad[now] && e[i].to != son[now])
dfs1(e[i].to);
}
int lca(int x, int y) {
while(top[x] != top[y]) {
if(deep[top[x]] < deep[top[y]]) swap(x, y);
x = dad[top[x]];
}
return deep[x] > deep[y] ? y : x;
}
int main () {
n = read();q = read();
for(int i = , u, v; i < n; i++) {
u = read();v = read();
add(u, v), add(v, u);
}
dfs();
dfs1();
for(int i = , a, b, c, d; i <= q; i++) {
a = read(), b = read(), c = read(), d = read();
int fir = max(deep[lca(a, b)], deep[lca(c, d)]);
int sec = max(max(deep[lca(a, c)], deep[lca(a, d)]), max(deep[lca(b, c)], deep[lca(b, d)]));
if(sec >= fir) puts("Y");
else puts("N");
}
return ;
}

谢谢收看, 祝身体健康!

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