RMQ POJ3264
RMQ
RMQ(Regional Maximize/Minmize Query)区间最值问题
一般的处理方法是ST表，即对于序列A的每个位置i处理出一个数组dp[i][j] 表示 [i……i+(2^j)-1]这个区间的最值（也就是i往后走2^j步）（倍增思想）
最小区间[i,i]，显然dp[i][0] = a[i]；
继续往上推，dp[i][j]可以拆分为两个更小的部分dp[i][j-1]和dp[i+(2^(j-1))][j-1]……
RMQ
void RMQ()
{
for(int j=1;j<=30;++j){
for(int i=1;i<=n;++i){
if(i+(1<
最大最小值询问
一段区间可以被两段2^k长度覆盖（可能有重合），所以就可以轻松地询问最大最小值辣
int qmax(int l,int r)
{
int k=(int)(log(r-l+1.0)/log(2.0));
return max(ma[l][k],ma[r-(1<
POJ3264
Balanced Lineup
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 64360 | Accepted: 30000 | |
| Case Time Limit: 2000MS | ||
Description
For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.
Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.
Input
Line 1: Two space-separated integers, N and Q.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i
Lines N+2..N+Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.
Output
Lines 1..Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
直接上代码辣
#include
#include
#include
using namespace std;
const int maxn=50020;
int n,q;
int ma[maxn<<1][30];
int mi[maxn<<1][30];
void RMQ()
{
for(int j=1;j<=30;++j){
for(int i=1;i<=n;++i){
if(i+(1<