20200723T3 小X的佛光

Description

Input

Output

Sample Input

3 3 1
1 2
2 3
1 2 3
1 1 3
3 1 3

Sample Output

1
1
3

Data Constraint

 solution

题目意思就是求两条路径的交点数

LCA求即可

有两个点是一条200000的链

函数没开inline会爆栈

以后函数都加inline吧。。。

mlg别再折磨我的code了。。。

放一下官方题解

 

code

#include<iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<queue>
#include<vector>
#include<stack>
#include<set>
#include<deque>
#include<map>
using namespace std;

template <typename T> void read(T &x) {
    x = 0; int f = 1; char c;
    for (c = getchar(); c < '0' || c > '9'; c = getchar()) if (c == '-') f = -f;
    for (; c >= '0' && c <= '9'; c = getchar()) x = 10 * x + c - '0' ;
    x *= f;
}
template <typename T> void write(T x){
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) { write(x); putchar('
'); }
template <typename T> void writesn(T x) { write(x); putchar(' '); }

#define int long long
#define inf 1234567890
#define next net
#define P 1000000007
#define N 400020
#define mid ((l+r)>>1)
#define lson (o<<1)
#define rson (o<<1|1)
#define R register

int n, q, num , A, B, C;
int cut, head[N ], next[N ], ver[N ];
int deep[N ], f[N ][25];
inline void add(int x,int y)
{
    ver[++cut] = y; next[cut] = head[x]; head[x] = cut;
}
inline void dfs(int x,int fa)
{
    deep[x] = deep[fa] + 1;
    f[x][0] = fa;
    for(R int i = head[x]; i; i = next[i])
    {
        int y = ver[i];
        if(y == fa )continue;
        dfs(y, x);
    }
}
inline void work()
{
    for(R int i = 1; i <= 20; i++)
        for(R int j = 1; j <= n; j++)
            f[j][i] = f[ f[j][i - 1] ][i - 1];
}
inline int lca(int x,int y)
{
    if(deep[x] < deep[y]) swap(x, y);
    for(R int i = 20; i >= 0; i--)
        if(deep[ f[x][i] ] >= deep[y]) x = f[x][i];
    if(x == y) return x;
    for(R 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];
}
inline int calc(int x,int y)
{
    return deep[x] + deep[y] - (deep[lca(x, y)] << 1) + 1;
}
signed main()
{
    read(n); read(q); read(num );
    for(R int i = 1, x, y; i < n; i++)
    {
        read(x); read(y);
        add(x,y); add(y,x);
    }
    dfs(1,0);
    work();
    while(q--)
    {
        read(A); read(B); read(C);
        writeln(calc(A, B) + calc(B, C) - calc(A, C) +1 >> 1);
    }
    return 0;
}