poj3264 Balanced Lineup
來源：程序員人生 發布時間：2015-05-29 08:30:19 閱讀次數：2416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
這道題可以用線段樹做，定義1個結構體，定義兩個變量high,low,其中high記錄這條線段中最大的點，low為最小的，每次更新到b[i].l==l&&b[i].r==r判斷這段最大值和最小值和原來所記錄的相比怎樣樣。
#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;
}
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