poj-3264-Balanced Lineup
poj 3264 Balanced Lineup
link: http://poj.org/problem?id=3264
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 48747 | Accepted: 22833 | |
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
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
Sample Input
6 3
1
7
3
4
2
5
1 5
4 6
2 2
Sample Output
6
3
0
Source
题解:
快速找到一个区间[a, b] 之间的最大值和最小值的差;
经典的RMQ问题。 利用Sparse Table算法, 动态规划求解。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
const int maxn = 50005; int n, m, num[maxn], dp1[maxn][18], dp2[maxn][18]; void BuildIndex(){
for(int i=0; i<n;++i){
dp1[i][0] = i;
dp2[i][0] = i;
}
for(int i=1; (1<<i)<=n; ++i){
for(int j=0; j+(1<<i)-1<n; ++j){
// find max
if(num[dp1[j][i-1]] > num[dp1[j+(1<<(i-1))][i-1]]){
dp1[j][i] = dp1[j][i-1];
}else{
dp1[j][i] = dp1[j+(1<<(i-1))][i-1];
} // find min
if(num[dp2[j][i-1]] < num[dp2[j+(1<<(i-1))][i-1]]){
dp2[j][i] = dp2[j][i-1];
}else{
dp2[j][i] = dp2[j+(1<<(i-1))][i-1];
}
}
}
} int FindMaxIndex(int start, int end){
int k = (int)((log((end - start + 1)*1.0))/log(2.0));
if(num[dp1[start][k]] > num[dp1[end-(1<<k)+1][k]]){
return dp1[start][k];
}else{
return dp1[end-(1<<k)+1][k];
}
}
int FindMinIndex(int start, int end){
int k = (int)((log((end - start + 1)*1.0))/log(2.0));
if(num[dp2[start][k]] > num[dp2[end-(1<<k)+1][k]]){
return dp2[end-(1<<k)+1][k];
}else{
return dp2[start][k];
}
} int main(){
freopen("in.txt", "r", stdin); int ans1, ans2, x, y;
while(scanf("%d %d", &n, &m) != EOF){
for(int i=0; i<n; ++i){
scanf("%d", &num[i]);
}
BuildIndex();
while(m--){
scanf("%d %d", &x, &y);
if(x > y){ swap(x, y); }
ans1 = FindMinIndex(x-1, y-1);
ans2 = FindMaxIndex(x-1, y-1);
printf("%d\n", (num[ans2] - num[ans1]) );
}
}
return 0;
}
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