poj_3264 Balanced Lineup 线段树+点修改/RMQ

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 35257		Accepted: 16586
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 andQ.

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
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

线段树写法：

运行结果：

#include<stdio.h>
#include<cstdio>
#include <cstring>
#include<iostream>
#include<algorithm>
#define inf -0x3f3f3f3f
#define INF 0x3f3f3f3f
#define mem0(a) memset(a,0,sizeof(a))
using namespace std;
const int  num = 50000+10;
int _max[num<<2],_min[num<<2];
void GetMax(int cur){
_max[cur] = max(_max[cur<<1],_max[cur<<1|1]);
}
void GetMin(int cur){
_min[cur] = min(_min[cur<<1],_min[cur<<1|1]);
}
void build(int l,int r,int cur){
if( l == r ){
scanf("%d",&_max[cur]);
_min[cur] = _max[cur];
return ;
}
int mid= (l + r )>>1;
build(l,mid,cur<<1);
build(mid+1,r,cur<<1|1);
GetMax(cur);
GetMin(cur);
}

int  query1(int ql,int qr,int l,int r,int cur){
int mid = ( l + r )>>1 ;
int ma =  inf;
if(ql <= l && qr >= r)  return _max[cur];
if(ql <= mid) ma = max( ma , query1(ql,qr,l,mid,cur<<1));
if(qr > mid ) ma = max( ma , query1(ql,qr,mid+1,r,cur<<1|1));
return ma ;
}
int  query2(int ql,int qr,int l,int r,int cur){
int mid = ( l + r )>>1 ;
int mi = INF;
if(ql <= l && qr >= r)  return _min[cur];
if(ql <= mid) mi = min(mi , query2(ql,qr,l,mid,cur<<1));
if(qr > mid ) mi = min(mi , query2(ql,qr,mid+1,r,cur<<1|1));
return mi ;
}
int main()
{
int N,M;
while(scanf("%d%d",&N,&M)!=EOF){
build(1,N,1);
while(M--){
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",query1(a,b,1,N,1)-query2(a,b,1,N,1));
}
}
return 0;
}

RMQ方法解:

有关RMQ算法讲解可以参考：RMQ (Range Minimum/Maximum Query)算法

运行结果：

 RMQ算法运行时间要比线段树少点，在南阳理工士兵杀敌3中，ＲＭＱ可以过，线段树就超时。。因此遇到类似问题建议用ＲＭＱ算法

#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;

const int N = 50010;//2^20=1048576>50010
int maxsum
[20], minsum
[20];

void RMQ(int num) //预处理->O(nlogn)
{
for(int j = 1; j < 20; ++j)
for(int i = 1; i <= num; ++i)
if(i + (1 << j) - 1 <= num)
{
maxsum[i][j] = max(maxsum[i][j - 1], maxsum[i + (1 << (j - 1))][j - 1]);
minsum[i][j] = min(minsum[i][j - 1], minsum[i + (1 << (j - 1))][j - 1]);
}
}

int main()
{
int num, query;
int src, des;
scanf("%d %d", &num, &query);
for(int i = 1; i <= num; ++i) //输入信息处理
{
scanf("%d", &maxsum[i][0]);
minsum[i][0] = maxsum[i][0];
}
RMQ(num);
while(query--) //O(1)查询
{
scanf("%d %d", &src, &des);
int k = (int)(log(des - src + 1.0) / log(2.0));
int maxres = max(maxsum[src][k], maxsum[des - (1 << k) + 1][k]);
int minres = min(minsum[src][k], minsum[des - (1 << k) + 1][k]);
printf("%d\n", maxres - minres);
}
return 0;
}