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| Time Limit: 5000MS | Memory Limit: 65536K | |
| Total Submissions: 44720 | Accepted: 20995 | |
| 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
Source
ST算法求区间内最值
1 #include<algorithm> 2 #include<iostream> 3 #include<cstdio> 4 #include<cstring> 5 #include<cmath> 6 using namespace std; 7 const int mxn=60000; 8 int n,Q; 9 int h[mxn]; 10 int fmx[mxn][20],fmi[mxn][20];//f[i][j]表示从i开始到i*（1<<j）范围内的目标值 11 int ST(int a,int b){//ST算法，倍增求区间最值 12 int i,j; 13 for(i=1;i<=n;i++){//自身 14 fmx[i][0]=fmi[i][0]=h[i]; 15 } 16 int k=(int)(log(n*1.0)/log(2.0));//k=以2为底n的对数 17 for(i=1;i<=k;i++){//第一层为次数 18 for(j=1;j<=n;j++){//第二层为范围 19 fmx[j][i]=fmx[j][i-1]; 20 if(j+(1<<(i-1)) <=n) 21 fmx[j][i]=max(fmx[j][i],fmx[j+(1<<(i-1))][i-1]); 22 fmi[j][i]=fmi[j][i-1]; 23 if(j+(1<<(i-1)) <=n) 24 fmi[j][i]=min(fmi[j][i],fmi[j+(1<<(i-1))][i-1]); 25 } 26 } 27 return 0; 28 } 29 int ansmx(int a,int b){//查找最大值 30 int k=(int)(log(b-a+1.0)/log(2.0)); 31 return max(fmx[a][k],fmx[b-(1<<k)+1][k]); 32 } 33 int ansmi(int a,int b){//查找最小值 34 int k=(int)(log(b-a+1.0)/log(2.0)); 35 return min(fmi[a][k],fmi[b-(1<<k)+1][k]); 36 } 37 int main(){ 38 scanf("%d%d",&n,&Q); 39 int i,j; 40 for(i=1;i<=n;i++){ 41 scanf("%d",&h[i]); 42 } 43 int a,b; 44 ST(1,n); 45 for(i=1;i<=Q;i++){ 46 scanf("%d%d",&a,&b); 47 printf("%d\n",ansmx(a,b)-ansmi(a,b));//输出询问区间内最大值和最小值的差 48 } 49 return 0; 50 }
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原文地址：http://www.cnblogs.com/SilverNebula/p/5638900.html
