POJ 3264 Balanced Lineup
2013-06-25 20:52
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[align=center]Balanced Lineup[/align]
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
Sample Output
Source
USACO 2007 January Silver
RMQ算法模版题。注意运算符的优先级 1 + 1<<2 和1 + (1<<2)的结果是不一样的
同样1<<2 + 1 和(1<<2)+1的结果也是不一样的。
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 27620 | Accepted: 12987 | |
Case Time Limit: 2000MS |
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
RMQ算法模版题。注意运算符的优先级 1 + 1<<2 和1 + (1<<2)的结果是不一样的
同样1<<2 + 1 和(1<<2)+1的结果也是不一样的。
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> #define N 51000 using namespace std; int Max [20],Min [20],a ,n,m; int main() { //freopen("data1.in","r",stdin); void RMQ_init(); void RMQ(int l,int r,int &res1,int &res2); while(scanf("%d %d",&n,&m)!=EOF) { for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } RMQ_init(); for(int i=1;i<=m;i++) { int x,y; scanf("%d %d",&x,&y); int res1,res2; RMQ(x,y,res1,res2); printf("%d\n",res2 - res1); } } return 0; } void RMQ_init() { for(int i=1;i<=n;i++) { Max[i][0] = a[i]; Min[i][0] = a[i]; } for(int j=1;(1<<j)<=n;j++) { for(int i=1;i +(1<<(j-1))<=n;i++) { Min[i][j] = min(Min[i][j-1],Min[i +(1<<(j-1))][j-1]); Max[i][j] = max(Max[i][j-1],Max[i +(1<<(j-1))][j-1]); } } } void RMQ(int l,int r,int &res1,int &res2) { int k = 0; while(1<<(k+1)<=(r - l + 1)) { k++; } res1 = min(Min[l][k],Min[r -(1<<k)+1][k]); res2 = max(Max[l][k],Max[r -(1<<k)+1][k]); }
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