poj3264 Balanced Lineup
来源：程序员人生 发布时间：2015-05-29 08:30:19 阅读次数：2232次
| Time Limit: 5000MS |
|
Memory Limit: 65536K |
| Total Submissions: 37683 |
|
Accepted: 17656 |
| 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
这道题可以用线段树做，定义1个结构体，定义两个变量high,low,其中high记录这条线段中最大的点，low为最小的，每次更新到b[i].l==l&&b[i].r==r判断这段最大值和最小值和原来所记录的相比怎样样。
#include<stdio.h>
#include<string.h>
#define inf 88888888
int max(int a,int b){
return a>b?a:b;
}
int min(int a,int b){
return a<b?a:b;
}
int low,high;
int a[50006];
struct node
{
int l,r,high,low;
}b[4*50006];
void build(int l,int r,int i)
{
int mid;
b[i].l=l;b[i].r=r;
if(l==r){
b[i].low=b[i].high=a[l];return;
}
mid=(l+r)/2;
build(l,mid,i*2);
build(mid+1,r,i*2+1);
b[i].high=max(b[i*2].high,b[i*2+1].high);
b[i].low=min(b[i*2].low,b[i*2+1].low);
}
void question(int l,int r,int i)
{
int mid;
if(b[i].l==l && b[i].r==r){
if(b[i].high>high)high=b[i].high;
if(b[i].low<low)low=b[i].low;
return;
}
mid=(b[i].l+b[i].r)/2;
if(r<=mid)question(l,r,i*2);
else if(l>mid)question(l,r,i*2+1);
else {
question(l,mid,i*2);question(mid+1,r,i*2+1);
}
}
int main()
{
int n,m,i,j,c,d;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=1;i<=n;i++){
scanf("%d",&a[i]);
}
build(1,n,1);
for(i=1;i<=m;i++){
scanf("%d%d",&c,&d);
low=inf;high=0;
question(c,d,1);
printf("%d
",high-low);
}
}
return 0;
}
生活不易，码农辛苦
如果您觉得本网站对您的学习有所帮助,可以手机扫描二维码进行捐赠
