标签:space getc out using name main class span its
题意:求树上两条路径有无祖先。
思路:
瞎搞\(LCA\)啊。。。
可惜我\(LCA\)打错了,我居然调了半小时...qwq
#include <bits/stdc++.h>
using namespace std;
const int maxn = 200010;
#define travel(i,x) for(int i = head[x];i;i=e[i].nxt)
inline int read () {
int q=0,f=1;char ch = getchar();
while(!isdigit(ch)){
if(ch==‘-‘)f=-1;ch=getchar();
}
while(isdigit(ch)){
q=q*10+ch-‘0‘;ch=getchar();
}
return q*f;
}
struct edge {
int to;
int nxt;
}e[maxn << 1];
int cnt;
int head[maxn << 1];
inline void add(int u,int v) {
e[++cnt].to = v;
e[cnt].nxt = head[u];
head[u] = cnt;
}
int f[maxn][30];
int dep[maxn << 1];
inline void dfs(int x,int fa) {
dep[x] = dep[fa] + 1;
travel(i,x) {
int y = e[i].to;
if(y != fa) {
f[y][0] = x;
for(int j = 1;j <= 20; ++j) {
f[y][j] = f[f[y][j - 1]][j - 1];
}
dfs(y,x);
}
}
}
inline int lca(int x,int y) {
if(dep[x] < dep[y]) swap(x,y);
for(int i = 20;i >= 0; --i) {
if(dep[f[x][i]] >= dep[y]) x = f[x][i];
}
if(x == y) return x;
for(int i = 20; i >= 0; --i) {
if(f[x][i] != f[y][i]) {
x = f[x][i];
y = f[y][i];
}
}
return f[x][0];
}
int n,m;
int main () {
freopen("inter.in","r",stdin);
freopen("inter.out","w",stdout);
n = read();
for(int i = 1;i < n; ++i) {
int x = read(),y = read();
add(x,y);add(y,x);
}
dfs(1,0);
m = read();
while(m--) {
int x = read(),y = read();
int LCA = lca(x,y);
int l = read(),r = read();
int _LCA = lca(l,r);
//cout<<"LCA:"<<LCA<<‘ ‘<<_LCA<<endl;
int Grand_LCA = lca(LCA,_LCA);
if(LCA == _LCA) {
puts("YES");
}
else if(Grand_LCA == _LCA && (lca(l,LCA) == LCA || lca(r,LCA) == LCA)) {
puts("YES");
}
else if(Grand_LCA == LCA && (lca(x,_LCA) == _LCA || lca(y,_LCA) == _LCA)) {
puts("YES");
}
else puts("NO");
}
return 0;
}
标签:space getc out using name main class span its
原文地址:https://www.cnblogs.com/akoasm/p/9607335.html