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B. Balanced Lineup

2024-07-11 21:34:54   222人阅读

B. Balanced Lineup

5000ms
5000ms
65536KB
 
64-bit integer IO format: %lld      Java class name: Main
 

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

解题：RMQ

 1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <vector> 6 #include <climits> 7 #include <algorithm> 8 #include <cmath> 9 #define LL long long10 using namespace std;11 int mn[50010][32],mx[50010][32],d[50010];12 int main(){13     int n,m,x,y,i,j;14     while(~scanf("%d %d",&n,&m)){15         for(i = 0; i < n; i++){16             scanf("%d",d+i);17         }18         memset(mn,0,sizeof(mn));19         memset(mx,0,sizeof(mx));20         for(i = n-1; i >= 0; i--){21             mn[i][0] = mx[i][0] = d[i];22             for(j = 1; i+(1<<j)-1 < n; j++){23                 mn[i][j] = min(mn[i][j-1],mn[i+(1<<(j-1))][j-1]);24                 mx[i][j] = max(mx[i][j-1],mx[i+(1<<(j-1))][j-1]);25             }26         }27         for(i = 0; i < m; i++){28             scanf("%d %d",&x,&y);29             if(x > y) swap(x,y);30             int r = y - x + 1;31             r = log2(r);32             int theMax,theMin;33             theMax = max(mx[x-1][r],mx[y-(1<<r)][r]);34             theMin = min(mn[x-1][r],mn[y-(1<<r)][r]);35             printf("%d\n",theMax-theMin);36         }37     }38     return 0;39 }
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