poj 3246 Balanced Lineup(线段树)
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 38942 | Accepted: 18247 | |
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
Lines 2..N+1: Line i+1 contains a single integer that is the height of cowi
Lines N+2..N+Q+1: Two integers A and B (1 ≤A ≤
B ≤ N), representing the range of cows from A toB inclusive.
Output
Sample Input
6 3
1
7
3
4
2
5
1 5
4 6
2 2
Sample Output
6
3
0
Source
题目大意:在一定区间内给定一些数。要求求出在某一区间内最大值和最小值的差。
线段树的题目。对于这道题目,既然是求最大值和最小值的差,那么必定要在区间里面存放最大值和最小值。同一时候这道题目仅仅是单纯的要求查询区间内的差值,不须要进行更新。
#include<stdio.h>
#include<string.h>
#define max(a,b) a>b?a:b
#define min(a,b) a<b? a:b
#define INF 99999999
#define N 50005
struct tree{
int l,r,maxi,mini;
int mid(){
return l+r>>1;
}
}tree[N<<2];
int ma=-INF,mi=INF;
void build(int l,int r,int root)
{
tree[root].l=l;
tree[root].r=r;
tree[root].maxi=-INF;
tree[root].mini=INF; //初始化最大最小值
if(l==r){ return;
}
int mid=l+r>>1;
build(l,mid,root<<1);
build(mid+1,r,root<<1|1);
}
void update(int i,int z,int root)
{ if(tree[root].l==tree[root].r){
tree[root].mini=tree[root].maxi=z;
return;
}
tree[root].maxi=max(tree[root].maxi,z);
tree[root].mini=min(tree[root].mini,z); //每次都更新最大和最小值
if(i<=tree[root].mid())update(i,z,root<<1); //这里将i下面的节点所有更新。 而i与mid 是有关系的。
else update(i,z,root<<1|1);
}
void Query(int l,int r,int root)
{
if(tree[root].mini>=mi&&tree[root].maxi<=ma)return;
if(l==tree[root].l&&r==tree[root].r){
mi=min(mi,tree[root].mini);
ma=max(ma,tree[root].maxi);
return;
}
int mid=tree[root].l+tree[root].r>>1;
if(r<=mid){
Query(l,r,root<<1);
}
else if(l>mid){
Query(l,r,root<<1|1);
}
else {
Query(l,mid,root<<1);
Query(mid+1,r,root<<1|1);
}
return ;
}
int main()
{
int n,Q,cow[200005],a,b;
int i,j,k;
while(scanf("%d%d",&n,&Q)!=EOF)
{
build(1,n,1);
for(i=1;i<=n;i++)
{
scanf("%d",&cow[i]);
update(i,cow[i],1); //对于第i个数字进行插入
} while(Q--)
{
scanf("%d%d",&a,&b);
ma=-INF;
mi=INF;
Query(a,b,1);
printf("%d\n",ma-mi);
}
}
return 0;
}
最新文章
- golang-web框架revel一个表单提交的总结
- Webform 上传图片加水印
- hdu 2043
- com组件的注册
- star
- arcgis gdb含下划线_和%的查询 by gisoracle
- 省常中模拟 Test3 Day2
- KNN及其改进算法的python实现
- 使用 Windows 窗体 TextBox 控件创建密码文本框
- WebSocket使用教程 - 带完整实例
- java_内存划分
- cf472C Design Tutorial: Make It Nondeterministic
- android开发之多线程实现方法概述
- live-server 介绍&;安装
- 基于create-react-app的再配置
- 自学Python之路-Python核心编程
- jquery 判断元素是否可见
- Facelets应用程序的生命周期
- 1. eclipse异常问题解决办法
- [转]C++ STL list的初始化、添加、遍历、插入、删除、查找、排序、释放