标签：des style blog http java color
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.
6 3 1 7 3 4 2 5 1 5 4 6 2 2
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\n",theMax-theMin); 36 } 37 } 38 return 0; 39 }
B. Balanced Lineup,布布扣,bubuko.com
标签：des style blog http java color
原文地址：http://www.cnblogs.com/crackpotisback/p/3841068.html
