POJ3264-Balanced Lineup

Balanced Lineup

Time Limit: 5000MS Memory Limit: 65536K
Total Submissions: 52035 Accepted: 24413
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
Source

USACO 2007 January Silver

题目大意：一段序列，求给定区间的极差
**解题思路：**RMQ，下标从1开始

vector版本（AC）：
Memory: 8872K Time: 4829MS

#include<iostream>#include<algorithm>#include<cstdlib>#include<vector>#include<cstring>#include<cstdio>using namespace std;const int MAXN=50005;int d[MAXN][20];int p[MAXN][20];vector<int> A;//O(nlogn)预处理void initRMQ(const vector<int>& A){    int n=A.size();    for(int i=0;i<n;i++) d[i][0]=A[i],p[i][0]=A[i];    for(int j=1;(1<<j)<=n;j++)        for(int i=0;i+(1<<j)-1<n;i++)        {            d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);            p[i][j]=max(p[i][j-1],p[i+(1<<(j-1))][j-1]);        }}//O(1)查询int rmqmin(int L,int R){    int k=0;    while((1<<(k+1))<=R-L+1) k++;    return min(d[L][k],d[R-(1<<k)+1][k]);}int rmqmax(int L,int R){    int k=0;    while((1<<(k+1))<=R-L+1) k++;    return max(p[L][k],p[R-(1<<k)+1][k]);}int main(){    ios::sync_with_stdio(false);    cin.tie(0);    int n,q;    int height;    while(cin>>n>>q)    {        A.clear();        A.push_back(0);        for(int i=0;i<n;i++)        {            cin>>height;            A.push_back(height);        }        initRMQ(A);        int L,R;        int Max,Min;        for(int i=0;i<q;i++)        {            cin>>L>>R;            Max=rmqmax(L,R);            Min=rmqmin(L,R);            cout<<Max-Min<<endl;        }    }    return 0;}

数组版本（AC）：
Memory: 8720K Time: 3485MS

#include<iostream>#include<algorithm>#include<cstdlib>#include<vector>#include<cstring>#include<cstdio>using namespace std;const int MAXN=50005;int d[MAXN][20];int p[MAXN][20];int A[MAXN];int n,q;//O(nlogn)预处理void initRMQ(const int A[]){    n++;    for(int i=0;i<n;i++) d[i][0]=A[i],p[i][0]=A[i];    for(int j=1;(1<<j)<=n;j++)        for(int i=0;i+(1<<j)-1<n;i++)        {            d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);            p[i][j]=max(p[i][j-1],p[i+(1<<(j-1))][j-1]);        }}//O(1)查询int rmqmin(int L,int R){    int k=0;    while((1<<(k+1))<=R-L+1) k++;    return min(d[L][k],d[R-(1<<k)+1][k]);}int rmqmax(int L,int R){    int k=0;    while((1<<(k+1))<=R-L+1) k++;    return max(p[L][k],p[R-(1<<k)+1][k]);}int main(){    while(scanf("%d%d",&n,&q)!=EOF)    {        A[0]=0;        for(int i=1;i<=n;i++)        {            scanf("%d",&A[i]);        }        initRMQ(A);        int L,R;        int Max,Min;        for(int i=0;i<q;i++)        {            scanf("%d%d",&L,&R);            Max=rmqmax(L,R);            Min=rmqmin(L,R);            printf("%d\n",Max-Min);        }    }    return 0;}

数组版本（TLE）：

#include<iostream>#include<algorithm>#include<cstdlib>#include<vector>#include<cstring>#include<cstdio>using namespace std;const int MAXN=50005;int d[MAXN][20];int p[MAXN][20];int A[MAXN];int n,q;//O(nlogn)预处理void initRMQ(const int A[]){    n++;    for(int i=0;i<n;i++) d[i][0]=A[i],p[i][0]=A[i];    for(int j=1;(1<<j)<=n;j++)        for(int i=0;i+(1<<j)-1<n;i++)        {            d[i][j]=min(d[i][j-1],d[i+(1<<(j-1))][j-1]);            p[i][j]=max(p[i][j-1],p[i+(1<<(j-1))][j-1]);        }}//O(1)查询int rmqmin(int L,int R){    int k=0;    while((1<<(k+1))<=R-L+1) k++;    return min(d[L][k],d[R-(1<<k)+1][k]);}int rmqmax(int L,int R){    int k=0;    while((1<<(k+1))<=R-L+1) k++;    return max(p[L][k],p[R-(1<<k)+1][k]);}int main(){    ios::sync_with_stdio(false);    cin.tie(0);    while(cin>>n>>q)    {        A[0]=0;        for(int i=1;i<=n;i++)        {            cin>>A[i];        }        initRMQ(A);        int L,R;        int Max,Min;        for(int i=0;i<q;i++)        {            cin>>L>>R;            Max=rmqmax(L,R);            Min=rmqmin(L,R);            cout<<Max-Min<<endl;        }    }    return 0;}