pku 3264 Balanced Lineup ( rmq or 线段树 )

Balanced Lineup

Time Limit: 5000MS Memory Limit: 65536KTotal Submissions: 27941 Accepted: 13120Case 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 31734251 54 62 2

Sample Output

630

Source

USACO 2007 January Silver

题意:给出n个高度,求一个区间的最大差值

题解:可以用2棵线段树,一棵保存最大值,一棵保存最小值,结果就是相减.
也可以用rmq做,也是一个dp区间保存最大值,另一个保存最小值.

//segememt tree

#include<stdio.h>#include<math.h>int tree[200005],tree2[200005],c[50005];int MAX(int a,int b){return a>b?a:b;}int MIN(int a,int b){return a<b?a:b;}void init(int l,int r,int pos){    int mid=(l+r)/2;    if(l==r){ tree[pos]=c[l]; return;}    init(l,mid,2*pos);    init(mid+1,r,2*pos+1);    tree[pos]=MAX(tree[2*pos],tree[2*pos+1]);}void init2(int l,int r,int pos){    int mid=(l+r)/2;    if(l==r){ tree2[pos]=c[l]; return;}    init2(l,mid,2*pos);    init2(mid+1,r,2*pos+1);    tree2[pos]=MIN(tree2[2*pos],tree2[2*pos+1]);}int ques(int l,int r,int pos,int templ,int tempr){    int mid=(l+r)/2;    if(templ<=l&&r<=tempr) return tree[pos];    if(templ>mid) return ques(mid+1,r,2*pos+1,templ,tempr);    else if(tempr<=mid) return ques(l,mid,2*pos,templ,tempr);    else return MAX(ques(mid+1,r,2*pos+1,mid+1,tempr),ques(l,mid,2*pos,templ,mid));}int ques2(int l,int r,int pos,int templ,int tempr){    int mid=(l+r)/2;    if(templ<=l&&r<=tempr) return tree2[pos];    if(templ>mid) return ques2(mid+1,r,2*pos+1,templ,tempr);    else if(tempr<=mid) return ques2(l,mid,2*pos,templ,tempr);    else return MIN(ques2(mid+1,r,2*pos+1,mid+1,tempr),ques2(l,mid,2*pos,templ,mid));}int main(){    int n,q,i,x,y;    while(scanf("%d%d",&n,&q)>0)    {        for(i=1;i<=n;i++) scanf("%d",c+i);        init(1,n,1);        init2(1,n,1);        for(i=0;i<q;i++)        {            scanf("%d%d",&x,&y);            printf("%d\n",ques(1,n,1,x,y)-ques2(1,n,1,x,y));        }    }    return 0;}//rmq#include<stdio.h>#include<math.h>int dp[50005][38],dp2[50005][38],c[50005];int MAX(int a,int b){return a>b?a:b;}int MIN(int a,int b){return a<b?a:b;}int main(){    int n,q,i,j,x,y,k;    while(scanf("%d%d",&n,&q)>0)    {        for(i=0;i<n;i++)        {            scanf("%d",c+i);            dp[i][0]=dp2[i][0]=c[i];        }        for(i=1;(1<<i)<=n;i++)        for(j=0;j+(1<<i)-1<n;j++)        dp[j][i]=MAX(dp[j][i-1],dp[j+(1<<(i-1))][i-1]);        for(i=1;(1<<i)<n;i++)        for(j=0;j+(1<<i)-1<n;j++)        dp2[j][i]=MIN(dp2[j][i-1],dp2[j+(1<<(i-1))][i-1]);        for(i=0;i<q;i++)        {            scanf("%d%d",&x,&y);            x--,y--;            k=(int)(log((y-x+1)*1.0)/log(2.0));            printf("%d\n",MAX(dp[x][k],dp[y-(1<<k)+1][k])-MIN(dp2[x][k],dp2[y-(1<<k)+1][k]));        }    }    return 0;}