loj10135. 「一本通 4.4 练习 2」祖孙询问

思路：

　　在LCA函数中如果将x与y放置到同一深度时，若x==y则x与y有祖宗关系且deep较小的为祖宗，否则无祖宗关系

#include<cstdio>
#include<iostream>
#include<vector>
using namespace std;
const int maxn = 40010;
inline void qread(int &x){
    x = 0;
    register int ch = getchar(), flag = 0;
    while(ch < '0' || ch > '9')    {
        if(ch == '-')    flag = 1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9'){
        x = 10 * x + ch - 48;
        ch = getchar();
    }
    if(flag)    x = -x;
}
int n, m, rt = 1;
vector<int> G[maxn];
vector<int> W[maxn];
int llog[maxn];
int deep[maxn];
int f[maxn][30];
int g[maxn];
inline void init(){
    qread(n);
    for(int i=1; i<=n; ++i){
        int x, y;
        qread(x), qread(y);
        if(y == -1){
            rt = x;
            continue;
        }
        G[x].push_back(y);
        G[y].push_back(x);
    }
    qread(m);
    deep[rt] = 1;
}
void dfs(int x){
    for(int i=0; i<G[x].size(); ++i)
        if(!deep[G[x][i]]){
            f[G[x][i]][0] = x;
            deep[G[x][i]] = deep[x] + 1;
            dfs(G[x][i]);
        }
}
inline void STtree(){
    for(int i=2; i<=maxn - 10; ++i)
        llog[i] = llog[i >> 1] + 1;
    for(int j = 1; j <= llog[maxn - 10]; ++j)
        for(int i=1; i<=maxn - 10; ++i)
            f[i][j] = f[f[i][j-1]][j-1];        
}
inline int LCA(int x, int y){
    if(x == y)     return x;
    if(deep[x] < deep[y])    swap(x, y);
    for(int j = llog[n]; j>=0; --j)
        if(deep[f[x][j]] >= deep[y])
            x = f[x][j];
    if(x == y)     return 1;
    for(int j = llog[n]; j>=0; --j)
        if(f[x][j] != f[y][j]){
            x = f[x][j];
            y = f[y][j];
        }
    return 2;            
}
int main(void){
    init();
    dfs(rt);
    STtree();
    while(m--){
        int x, y;
        qread(x), qread(y);
        int z = LCA(x, y);
        if(z == 2)
            printf("0\n");
        else
            printf("%d\n", (deep[x] > deep[y]) + 1);
    }
}