POJ 3264 Balanced Lineup(RMQ)
2016-08-13 16:03
441 查看
#include<math.h> #include<stdio.h> #include<string.h> #include<algorithm> using namespace std; const int maxn=5e4+5; int maps[maxn],n,q; int dpmax[maxn][20],dpmin[maxn][20]; void init_rmq() { for(int i=1; i<=n; i++) dpmax[i][0]=dpmin[i][0]=maps[i]; for(int j=1; (1<<j)<=n; j++) { for(int i=1; i+(1<<j)-1<=n; i++) { dpmax[i][j]=max(dpmax[i][j-1],dpmax[i+(1<<(j-1))][j-1]); dpmin[i][j]=min(dpmin[i][j-1],dpmin[i+(1<<(j-1))][j-1]); } } } int get_ans(int a,int b) { int ans1,ans2,k=(int)((double)(log(b-a+1)*1.0)/log(2.0)); ans1=max(dpmax[a][k],dpmax[b-(1<<k)+1][k]); ans2=min(dpmin[a][k],dpmin[b-(1<<k)+1][k]); return ans1-ans2; } int main() { while(~scanf("%d%d",&n,&q)) { for(int i=1; i<=n; i++) { scanf("%d",&maps[i]); } init_rmq(); for(int i=1; i<=q; i++) { int a,b; scanf("%d%d",&a,&b); printf("%d\n",get_ans(a,b)); } } }
Balanced Lineup
Time Limit: 5000MS | Memory Limit: 65536K | |
Total Submissions: 46898 | Accepted: 21994 | |
Case Time Limit: 2000MS |
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
Source
USACO 2007 January Silver
相关文章推荐
- poj 3264 Balanced Lineup(RMQ裸题)
- POJ 3264 Balanced Lineup 线段树 RMQ
- POJ 3264 Balanced Lineup(线段树、RMQ)
- poj 3264 Balanced Lineup(RMQ && 线段树)
- POJ 3264 Balanced Lineup(RMQ)
- POJ 3264 Balanced Lineup (RMQ)
- poj 3264 --Balanced Lineup (RMQ或线段树)
- POJ 3264 Balanced Lineup (RMQ分析)
- POJ 3264 Balanced Lineup(简单RMQ)
- POJ 题目3264 Balanced Lineup(RMQ)
- POJ 3264 Balanced Lineup[RMQ入门题]
- POJ3264——Balanced Lineup 倍增RMQ裸题
- |poj 3264|RMQ|Balanced Lineup
- (简单) POJ 3264 Balanced Lineup,RMQ。
- POJ 3264 Balanced Lineup(RMQ)
- 【POJ】3264 - Balanced Lineup(RMQ - ST算法 || 线段树)
- POJ 3264 Balanced Lineup [RMQ]
- POJ 3264 Balanced Lineup(RMQ)
- POJ---3264-Balanced Lineup (RMQ)
- POJ 3264 Balanced Lineup (RMQ)