标签：min splay 最小值 sed include enter 2.0 cli cat
| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 49061 | Accepted: 22975 | |
| Case Time Limit: 2000MS | ||
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
Output
Sample Input
6 3 1 7 3 4 2 5 1 5 4 6 2 2
Sample Output
6 3 0
题目大意：给出一个数字序列，对于每次提问，输出区间（i,j）内的最大值数与最小值数的差。RMQ模板题。
#include<cstdio> #include<iostream> #include<cstring> #include<algorithm> #include<cmath> #define N 50010 #define ll long long using namespace std; int f[N][20],fi[N][20],a[N]; int n,m; void RMQ() { for (int i=1;i<=n;i++) f[i][0]=a[i],fi[i][0]=a[i]; for (int i=1;i<=floor(log(n)/log(2));i++) for (int j=1;j<=n+1-(1<<i);j++) { f[j][i]=max(f[j][i-1],f[j+(1<<i-1)][i-1]);//区间最大值 fi[j][i]=min(fi[j][i-1],fi[j+(1<<i-1)][i-1]);//区间最小值 } } int main() { while (scanf("%d%d",&n,&m)!=EOF) { for (int i=1;i<=n;i++) scanf("%d",&a[i]); RMQ(); for (int i=1,x,y;i<=m;i++) { scanf("%d%d",&x,&y); int k; k=(int) (log(y-x+1.0)/log(2.0)); //如果k=（log（y-x+1）/log（2）），那么当log（8）/log（2）时，k=2 int ans1=max(f[x][k],f[y-(1<<k)+1][k]),ans2=min(fi[x][k],fi[y-(1<<k)+1][k]); printf("%d\n",ans1-ans2); } } return 0; }
标签：min splay 最小值 sed include enter 2.0 cli cat
原文地址：http://www.cnblogs.com/sjymj/p/6065223.html
