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  • B. Balanced Lineup

    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 3
    1
    7
    3
    4
    2
    5
    1 5
    4 6
    2 2
     

    Sample Output

    6
    3
    0

    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 long
    10 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
    ",theMax-theMin);
    36         }
    37     }
    38     return 0;
    39 }
    View Code
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  • 原文地址:https://www.cnblogs.com/crackpotisback/p/3841068.html
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