转载注明出处： http://blog.csdn.net/u011400953
POJ 3264 Balanced Lineup 线段树求区间最大最小（普通线段树，ZKW线段树）
Balanced Lineup
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 30604 | Accepted: 14431 | |
| 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.
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
Source
USACO 2007 January Silver
【题目大意】：
给出一串的数字，然后给出一个区间a b，输出从a到b的最大的数和最小的数的差
【分析】：
用线段树，代码给出两种线段树：普通线段树和ZKW线段树
【代码1，普通线段树】：
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define MAX 50001
#define MAXP 250001
#define IMAX 214748364
int N,Q,a[MAX],value[MAXP],min1[MAXP],max1[MAXP],ans1,ans2;
void build(int node,int left,int right)
{
if(left==right)
{
min1[node]=a[left];
max1[node]=a[left];
}
else
{
int middle=(left+right)/2;
build(2*node,left,middle);
build(2*node+1,middle+1,right);
min1[node]=min(min1[2*node],min1[2*node+1]);
max1[node]=max(max1[2*node],max1[2*node+1]);
}
}
void find(int node,int left,int right,int L,int R)
{
if(L<=left && right<=R)
{
//printf("node:%d\n",node);
ans1=min(min1[node],ans1);
ans2=max(max1[node],ans2);
}
else
{
int middle=(left+right)/2;
if(L<=middle)
find(2*node,left,middle,L,R);
if(R>middle)
find(2*node+1,middle+1,right,L,R);
}
//maintain(node,left,right);
}
int main()
{
//freopen("input.in","r",stdin);
//freopen("output.out","w",stdout);
scanf("%d%d",&N,&Q);
for(int i=1;i<=N;i++)
scanf("%d",&a[i]);
build(1,1,N);
/*for(int i=1;i<=4*N;i++)
printf("min[%d]=%d\n max[%d]=%d\n",i,min1[i],i,max1[i]);*/
for(int i=1;i<=Q;i++)
{
int A,B;
ans1=IMAX;
ans2=-1;
scanf("%d%d",&A,&B);
find(1,1,N,A,B);
//printf("%d %d\n",ans1,ans2);
printf("%d\n",ans2-ans1);
}
//system("pause");
return 0;
}
【代码2，ZKW线段树】：
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define IMAX 21474836
#define MAX 50001
int N,Q,M,min1[4*MAX],max1[4*MAX];
int getmax(int left,int right)
{
int ansnow=-1;
for(left+=M-1,right+=M+1;left^right^1;left/=2,right/=2)
{
if(~left&1) ansnow=max(ansnow,max1[left^1]);
if(right&1) ansnow=max(ansnow,max1[right^1]);
}
return ansnow;
}
int getmin(int left,int right)
{
int ansnow=IMAX;
for(left+=M-1,right+=M+1;left^right^1;left/=2,right/=2)
{
if(~left&1) ansnow=min(ansnow,min1[left^1]);
if(right&1) ansnow=min(ansnow,min1[right^1]);
}
return ansnow;
}
int main()
{
//freopen("input.in","r",stdin);
//freopen("output.out","w",stdout);
scanf("%d%d",&N,&Q);
for(M=1;M<=N+1;M*=2);
for(int i=1;i<=N;i++)
{
scanf("%d",&min1[i]);
min1[i+M]=max1[i]=max1[i+M]=min1[i];
}
for(int i=M;i>=1;i--)
{
min1[i]=min(min1[2*i],min1[2*i+1]);
max1[i]=max(max1[2*i],max1[2*i+1]);
}
for(int i=1;i<=Q;i++)
{
int A,B;
scanf("%d%d",&A,&B);
printf("%d\n",getmax(A,B)-getmin(A,B));
}
//system("pause");
return 0;
}