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Give you a sequence of n numbers, and a number k you should find the max length of Good subsequence. Good subsequence is a continuous subsequence of the given sequence and its maximum value - minimum value<=k. For example n=5, k=2, the sequence ={5, 4, 2, 3, 1}. The answer is 3, the good subsequence are {4, 2, 3} or {2, 3, 1}.
There are several test cases.
Each test case contains two line. the first line are two numbers
indicates n and k (1<=n<=10,000, 1<=k<=1,000,000,000). The
second line give the sequence of n numbers a[i] (1<=i<=n,
1<=a[i]<=1,000,000,000).
The input will finish with the end of file.
For each the case, output one integer indicates the answer.
5 2
5 4 2 3 1
1 1
1
3 1
题意:在区间内找一段长度最大的子区间满足子区间的最大值 - 子区间的最小值 <=k ，输出最大值。
题解:尺取法扫一遍就可以得到了,区间最大最小用RMQ或者线段树都可以。
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; #define N 10010 int a[N]; int max_dp[N][20]; int min_dp[N][20]; void init_MAX_RMQ(int n){ for(int i=1;i<=n;i++) max_dp[i][0]=a[i]; for(int j=1;(1<<j)<=n;j++){ for(int i=1;i<=n-(1<<j)+1;i++){ max_dp[i][j] = max(max_dp[i][j-1],max_dp[i+(1<<(j-1))][j-1]); } } } int MAX_RMQ(int a,int b){ int k = (int)(log(b-a+1.0)/log(2.0)); return max(max_dp[a][k],max_dp[b-(1<<k)+1][k]); } void init_MIN_RMQ(int n){ for(int i=1;i<=n;i++) min_dp[i][0]=a[i]; for(int j=1;(1<<j)<=n;j++){ for(int i=1;i<=n-(1<<j)+1;i++){ min_dp[i][j] = min(min_dp[i][j-1],min_dp[i+(1<<(j-1))][j-1]); } } } int MIN_RMQ(int a,int b){ int k = (int)(log(b-a+1.0)/log(2.0)); return min(min_dp[a][k],min_dp[b-(1<<k)+1][k]); } int main() { int n,k; while(scanf("%d%d",&n,&k)!=EOF){ for(int i=1;i<=n;i++){ scanf("%d",&a[i]); } init_MAX_RMQ(n); init_MIN_RMQ(n); int MAX = -1; int maxv =a[1], minv = a[1]; int l = 1,r = 1; while(l<=n){ while(r<=n){ maxv = MAX_RMQ(l,r); minv = MIN_RMQ(l,r); if(maxv-minv>k) break; MAX = max(r-l+1,MAX); r++; } l++; maxv = MAX_RMQ(l,r); minv = MIN_RMQ(l,r); } printf("%d\n",MAX); } return 0; }
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原文地址：http://www.cnblogs.com/liyinggang/p/5793036.html
