Balanced lineup
Time limit:5000 Ms |
|
Memory limit:65536 K |
Total submissions:33094 |
|
Accepted:15552 |
Case time limit:2000 ms |
Description
For the daily milking, Farmer John'sNCows (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 shocould not differ too much in height.
Farmer John has made a listQ(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,
NAnd
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
AAnd
B(1 ≤
A≤
B≤
N), Representing the range of cows from
ATo
BIntrusive.
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
We plan to use two methods. By the way, the differences between rmq and line tree are different. They all said that rmq is better than a line segment tree, and I found that the time is not much different, even though it is not optimized.
The AC code is as follows:
Line Segment tree!
// 2250 Ms 2404 K # include <iostream> # include <cstring> # include <cstdio> # include <algorithm> # define M 50010 # define INF 100000000 using namespace std; struct H {int L, R, Maxx, Minn;} trees [4 * m]; int n, m; int num [m]; void build_trees (int jd, int l, int R) {trees [JD]. L = L; trees [JD]. R = r; If (L = r) {trees [JD]. maxx = num [l]; trees [JD]. minn = num [l]; return;} int mid = (L + r)/2; build_trees (JD * 2, L, mid); build_trees (JD * 2 + 1, Mid + 1, R); trees [JD]. maxx = max (trees [JD * 2]. maxx, trees [JD * 2 + 1]. maxx); trees [JD]. minn = min (trees [JD * 2]. minn, trees [JD * 2 + 1]. minn);} int query_max (int jd, int L, int R) {int ans = 0; If (L <= trees [JD]. L & R> = trees [JD]. r) return trees [JD]. maxx; int mid = (trees [JD]. L + trees [JD]. r)/2; If (L <= mid) ans = max (ANS, query_max (JD * 2, L, R); If (r> mid) ans = max (ANS, query_max (JD * 2 + 1, L, R); Return ans;} int query_min (int jd, int l, Int R) {int ans = inf; If (L <= trees [JD]. L & R> = trees [JD]. r) return trees [JD]. minn; int mid = (trees [JD]. L + trees [JD]. r)/2; If (L <= mid) ans = min (ANS, query_min (JD * 2, L, R); If (r> mid) ans = min (ANS, query_min (JD * 2 + 1, L, R); Return ans;} int main () {int I, j; int A, B; while (~ Scanf ("% d", & N, & M) {memset (Num, 0, sizeof num); for (I = 1; I <= N; I ++) scanf ("% d", & num [I]); build_trees (1, 1, n); for (I = 1; I <= m; I ++) {scanf ("% d", & A, & B); printf ("% d \ n", query_max (1, a, B)-query_min (1, a, B) ;}} return 0 ;}
Rmq !!!
///RMQ 1813MS 12100K#include<iostream>#include<cstring>#include<cstdio>#include<algorithm>#define M 50010#define inf 100000000using namespace std;int n,m;int num[M];int dp1[M][30],dp2[M][30];void RMQ_min(){ int i,j; memset(dp1,0,sizeof dp1); for(i=1;i<=n;i++) dp1[i][0]=num[i]; for(j=1;1<<j<=n;j++) for(i=1;i+(1<<j)-1<=n;i++) dp1[i][j]=min(dp1[i][j-1],dp1[i+(1<<(j-1))][j-1]);}void RMQ_max(){ int i,j; memset(dp2,0,sizeof dp2); for(i=1;i<=n;i++) dp2[i][0]=num[i]; for(j=1;1<<j<=n;j++) for(i=1;i+(1<<j)-1<=n;i++) dp2[i][j]=max(dp2[i][j-1],dp2[i+(1<<(j-1))][j-1]);}int rmq_min(int l,int r){ int i,j; int k=0; while(1<<(k+1)<=r-l+1) k++; return min(dp1[l][k],dp1[r-(1<<k)+1][k]);}int rmq_max(int l,int r){ int i,j; int k=0; while(1<<(k+1)<=r-l+1) k++; return max(dp2[l][k],dp2[r-(1<<k)+1][k]);}int main(){ int i,j; int a,b; while(~scanf("%d%d",&n,&m)) { for(i=1;i<=n;i++) scanf("%d",&num[i]); RMQ_min(); RMQ_max(); for(i=1;i<=m;i++) { scanf("%d%d",&a,&b); printf("%d\n",rmq_max(a,b)-rmq_min(a,b)); } } return 0;}