LCA与RMQ

参见topcoder的算法教程，Range Minimum Query and Lowest Common Ancestor

 

JOJ 一个题目：

2408 Beautiful girl， 题意：一颗树，给定三个顶点A、B、C，判断A与B之间的路径是否可以经过C

代码：

/*
 * LCA->RMQ
 */
#include<cstdio>
#include<vector>
#include<cmath>
using namespace std;
#define MY_MAX 50010

vector<int> graph[MY_MAX];

void read(int n){
    int i, u, v;
    for(i=0; i<n; ++i)
        if(!graph[i].empty())
            graph[i].clear();
    for(i=1; i<n; ++i){
        scanf("%d %d", &u, &v);
        graph[u].push_back(v);
        graph[v].push_back(u);
    }
}

int E[MY_MAX*2];//E[i] is the label of i-th visited node in the tour
int L[MY_MAX*2];//L[i] is the level of node E[i]
int H[MY_MAX];    //the index of the first occurrence of node i in E

bool visited[MY_MAX];
void DFS(int u, int d, int &k){
    visited[u] = true;
    E[k] = u;
    L[k] = d;
    H[u] = k;
    ++k;
    int v;
    for(size_t i=0; i<graph[u].size(); ++i){
        v = graph[u][i];
        if(!visited[v]){
            DFS(v, d+1, k);
            E[k] = u;
            L[k] = d;
            ++k;
        }
    }
}

int M[MY_MAX*2][20]; //存储的索引
void preprocess(int a[], int n){
    int i, j;
    //initialize for the intervals with length 1
    for(i=0; i<n; ++i)
        M[i][0] = i;
    //compute values from smaller to bigger intervals
    for(j=1; 1<<j <= n; ++j){
        for(i=0; i + (1<<j) - 1 < n; ++i)
            if(a[M[i][j - 1]] < a[M[i + (1<<(j-1))][j-1]])
                M[i][j] = M[i][j-1];
            else
                M[i][j] = M[i + (1<<(j-1))][j-1];
    }
}

int RMQ(int a[], int i, int j){
    if(i > j)
        swap(i, j);
    int k;
    k = int(log(j-i+1) / log(2.0));
    if(a[M[i][k]] < a[M[j - (1<<k) + 1][k]])
        return M[i][k];
    else
        return M[j - (1<<k) + 1][k];
}

int main(){
//    freopen("in", "r", stdin);
    int n, m, k, case_num = 1;
    while(scanf("%d", &n) != EOF){
        read(n);
        for(k=0; k<n; ++k)
            visited[k] = false;
        k = 0;
        DFS(0, 0, k);

        preprocess(L, 2*n-1);

        int a, b, c;
        int lab, lac, lbc;
        if(case_num != 1)
            printf("\n");
        printf("Case %d:\n", case_num++);

        scanf("%d", &m);
        for(k=0; k<m; ++k){
            scanf("%d %d %d", &a, &b, &c);
            lab = E[RMQ(L, H[a], H[b])];
            lac = E[RMQ(L, H[a], H[c])];
            lbc = E[RMQ(L, H[b], H[c])];
            if(a == c || b == c || (lab == lac && lbc == c)
                    || (lab == lbc && lac == c))
                printf("Yes\n");
            else
                printf("No\n");
        }
    }
}