「BZOJ1636」[Usaco2007 Jan] Balanced Lineup
Description
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
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Input
* 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.
to a reply and indicates the difference in height between the
tallest and shortest cow in the range.
Sample Output
6
3
0
3
0
题解
裸RMQ
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 |
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; inline int read() { int x=0;char ch=getchar(); while(ch<'0'||ch>'9')ch=getchar(); while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x; } int n,q; int s[16],mx[50005][16],mn[50005][16]; void pre() { int x=log(n)/log(2); for(int i=1;i<=x;i++) for(int j=1;j<=n;j++) { mx[j][i]=mx[j][i-1]; mn[j][i]=mn[j][i-1]; if(j+s[i-1]<=n)mx[j][i]=max(mx[j][i],mx[j+s[i-1]][i-1]); if(j+s[i-1]<=n)mn[j][i]=min(mn[j][i],mn[j+s[i-1]][i-1]); } } inline int rmq(int a,int b) { int x=log(b-a+1)/log(2); int t1=max(mx[a][x],mx[b-s[x]+1][x]); int t2=min(mn[a][x],mn[b-s[x]+1][x]); return t1-t2; } int main() { n=read(),q=read(); s[0]=1; for(int i=1;i<=15;i++) s[i]=s[i-1]*2; for(int i=1;i<=n;i++) mx[i][0]=mn[i][0]=read(); pre(); while(q--) { int x=read(),y=read(); printf("%d\n",rmq(x,y)); } return 0; } |
计数器 9612836
OI 课件=>
OI 课件=>
inline int read()
{
int x=0;char ch=getchar();
while(ch’9′)ch=getchar();
while(ch>=’0’&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x;
}求解释
读入优化……
已经知道了谢谢