标签：des style http color os io strong for
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 34488 | Accepted: 16203 | |
| 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
Source
就是套模板啦，算法训练指南第198页有详解。
AC代码：
#include<stdio.h>
#include<iostream>
#include<algorithm>
using namespace std;
int n,q;
int a[50010];
int d1[50010][110],d2[50010][110];
void RMQ_init(){
for(int i=1;i<=n;i++)
d1[i][0]=d2[i][0]=a[i];
for(int j=1;(1<<j)<=n;j++)
for(int i=1;i+(1<<j)-1<=n;i++){
d1[i][j]=max(d1[i][j-1],d1[i+(1<<(j-1))][j-1]);
d2[i][j]=min(d2[i][j-1],d2[i+(1<<(j-1))][j-1]);
}
}
int RMQ1(int x,int y){
int k=0;
while((1<<(k+1)<=y-x+1)) k++;
return max(d1[x][k],d1[y-(1<<k)+1][k]);
}
int RMQ2(int x,int y){
int k=0;
while(((1<<(k+1))<=y-x+1)) k++;
return min(d2[x][k],d2[y-(1<<k)+1][k]);
}
int main(){
while(scanf("%d%d",&n,&q)!=EOF){
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
RMQ_init();
while(q--){
int x,y;
scanf("%d%d",&x,&y);
printf("%d\n",RMQ1(x,y)-RMQ2(x,y));
}
}
return 0;
}
poj 3264(模板RMQ),布布扣,bubuko.com
标签：des style http color os io strong for
原文地址：http://blog.csdn.net/my_acm/article/details/38656287
