Greatest Parents Ligh OJ1128

其实倍增可以”缩”

题意:

一棵树的点权会给你,每一次询问一个x,y 问在x的祖先之中(包括x) 权值 >=y的而且是离x最远的点的编号!

思路

核心的什么也不说了,只是希望说一说这个:
每一次倍增,dp[i][j]~dp[i][0]其实就是相当于是一个二分搜索的过程呢!
比如dp[i][3]会符合答案,那么i就会自动”跳”到 i+2^3的地方,继续在(i+2^3)的地方2^2
一直下去,于是就有了答案!

代码:

#pragma GCC optimize(3)#include <bits/stdc++.h>#define N 100005#define M 20using namespace std;int T, n, m, x, y, val[N]={1}, fa[N][M], dp[N][M];vector<int> G[N];void DFS(int now){    for (int i = 0; i < G[now].size(); i++)    {        int v = G[now][i];        fa[v][0] = now, dp[v][0] = val[v];        for (int j = 1; j < M; j++)            fa[v][j] = fa[fa[v][j - 1]][j - 1];        for (int j = 1; j < M; j++)            dp[v][j] = max(dp[v][j - 1], dp[fa[v][j - 1]][j - 1]);        DFS(v);    }}inline int solve(int x, int y){    for (int i = M-1; i >= 0; i--)        if (dp[fa[x][i]][i] >= y)            x = fa[x][i];    return x;}int main(){    scanf("%d", &T);    for (int loc = 1; loc <= T; loc++)    {        scanf("%d%d", &n, &m);        for (int i = 0; i <= n; i++)            G[i].clear(), fa[0][i] = 0, dp[0][i] = 1;        for (int i = 1; i < n; i++)        {            scanf("%d%d", &x, &val[i]);            G[x].push_back(i);        }        DFS(0);        printf("Case %d:\n", loc);        for (int i = 1; i <= m; i++)        {            scanf("%d%d", &x, &y);            printf("%d\n", solve(x, y));        }    }    return 0;}