标签：acm
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 36215 | Accepted: 16954 | |
| 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
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
RMQ问题 第一发ST算法。。代码好挫
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define MAXN 55555
using namespace std;
int prelog2[MAXN],stmax[MAXN][35],stmin[MAXN][35];
int n,a[MAXN];
void init(){
prelog2[1]=0;
for(int i=2;i<=n;i++){
prelog2[i]=prelog2[i-1];
if((1<<prelog2[i]+1)==i){
prelog2[i]++;
}
}
for(int i=n;i>=1;i--){
stmax[i][0]=stmin[i][0]=a[i];
for(int j=1;(i+(1<<j)-1)<=n;j++){
stmax[i][j]=max(stmax[i][j-1],stmax[i+(1<<j-1)][j-1]);
stmin[i][j]=min(stmin[i][j-1],stmin[i+(1<<j-1)][j-1]);
}
}
}
int getmax(int l,int r){
int len=r-l+1;
return max(stmax[l][prelog2[len]],stmax[r-(1<<prelog2[len])+1][prelog2[len]]);
}
int getmin(int l,int r){
int len=r-l+1;
return min(stmin[l][prelog2[len]],stmin[r-(1<<prelog2[len])+1][prelog2[len]]);
}
int main(){
int q;
scanf("%d %d",&n,&q);
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
init();
while(q--){
int l,r;
scanf("%d %d",&l,&r);
printf("%d\n",getmax(l,r)-getmin(l,r));
}
return 0;
}
标签：acm
原文地址：http://blog.csdn.net/qq_16843991/article/details/41962907
