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里

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 = 100100;
int n, q, head[N], cnt, dad[N], top[N], size[N], deep[N], son[N];
struct edge{
    int next, to;
}e[N << 1];
int read() {
    int s = 0, w = 1;
    char ch = getchar();
    while(!isdigit(ch)) {if(ch == '-') w = -1;ch = getchar();}
    while(isdigit(ch)) {s = s * 10 + ch - '0';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] = 1;
    deep[now] = deep[dad[now]] + 1;
    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 = 1, u, v; i < n; i++) {
        u = read();v = read();
        add(u, v), add(v, u);
    } 
    dfs(1);
    dfs1(1);
    for(int i = 1, 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 0;
}

谢谢收看， 祝身体健康！

标签：include,int,题解,top,deep,son,suger,p3398,now
来源： https://www.cnblogs.com/yanxiujie/p/11439032.html