题目链接： http://acm.csu.edu.cn/csuoj/problemset/problem?pid=1553
Description
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}.
Input
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.
Output
For each the case, output one integer indicates the answer.
Sample Input
5 2 5 4 2 3 1 1 1 1
Sample Output
3 1
题解：
之前学了RMQ，线段树， 树状数组，但是一直不知道他们在哪里能派上用场。通过这题，终于找到他们的用武之地了：区间查询最大最小值。
解决了查询区间最大最小值的问题，剩下的就是二分了。这里是二分长度。
RMQ：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <string> 6 #include <vector> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <sstream> 11 #include <algorithm> 12 using namespace std; 13 #define pb push_back 14 #define mp make_pair 15 #define ms(a, b) memset((a), (b), sizeof(a)) 16 //#define LOCAL 17 #define eps 0.0000001 18 #define LNF (1<<60) 19 typedef long long LL; 20 const int inf = 0x3f3f3f3f; 21 const int maxn = 10000+10; 22 const int mod = 1e9+7; 23 24 LL st_max[maxn][16], st_min[maxn][16]; 25 LL a[maxn]; 26 27 void RMQ_init(int n) 28 { 29 for(int i = 0; i<n; i++) 30 { 31 st_min[i][0] = a[i]; 32 st_max[i][0] = a[i]; 33 } 34 35 for(int j = 1; (1<<j)<=n; j++)//枚举长度 36 for(int i = 0; i+(1<<j)-1<n; i++)//枚举起点 37 { 38 st_min[i][j] = min(st_min[i][j-1],st_min[i+(1<<(j-1))][j-1]); 39 st_max[i][j] = max(st_max[i][j-1],st_max[i+(1<<(j-1))][j-1]); 40 } 41 } 42 43 LL RMQ(int l, int r)//查询 44 { 45 int k = 0; 46 while((1<<(k+1))<=r-l+1) 47 k++; 48 return max(st_max[l][k],st_max[r-(1<<k)+1][k]) - min(st_min[l][k],st_min[r-(1<<k)+1][k]); 49 } 50 51 int test(int len, int n, int k) 52 { 53 for(int i = len-1; i<n; i++) 54 if(RMQ(i-len+1, i)<=1LL*k) 55 return 1; 56 57 return 0; 58 } 59 60 int main() 61 { 62 #ifdef LOCAL 63 freopen("input.txt", "r", stdin); 64 // freopen("output.txt", "w", stdout); 65 #endif // LOCAL 66 67 int n, k; 68 while(~scanf("%d%d", &n, &k)) 69 { 70 for(int i=0;i<n;i++) 71 scanf("%lld", &a[i]); 72 73 ms(st_max, 0); 74 ms(st_min, 0); 75 RMQ_init(n); 76 77 int l = 1, r = n; 78 while(l<=r) 79 { 80 int mid = (l+r)/2; 81 if(test(mid, n, k)) 82 l = mid+1; 83 else 84 r = mid-1; 85 } 86 printf("%d ", r); 87 } 88 return 0; 89 }
线段树：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cstdlib> 5 #include <string> 6 #include <vector> 7 #include <map> 8 #include <set> 9 #include <queue> 10 #include <sstream> 11 #include <algorithm> 12 using namespace std; 13 #define pb push_back 14 #define mp make_pair 15 #define ms(a, b) memset((a), (b), sizeof(a)) 16 //#define LOCAL 17 #define eps 0.0000001 18 #define LNF 1000000000000 19 typedef long long LL; 20 const int inf = 0x3f3f3f3f; 21 const int maxn = 10000+10; 22 const int mod = 1e9+7; 23 24 LL st_max[4*maxn], st_min[4*maxn]; 25 LL a[maxn]; 26 27 void build(int rt, int l, int r) 28 { 29 if(l==r) 30 { 31 st_max[rt] = a[r]; 32 st_min[rt] = a[r]; 33 return; 34 } 35 36 int mid = (l+r)>>1; 37 build(rt*2,l,mid); 38 build(rt*2+1,mid+1,r); 39 st_max[rt] = max(st_max[rt*2], st_max[rt*2+1]); 40 st_min[rt] = min(st_min[rt*2], st_min[rt*2+1]); 41 } 42 43 //由于最大最小值都要查询，而return只能返回一个，所以用ma和mi记录最小值 44 LL query(int rt, int l, int r, int x, int y, LL &ma, LL &mi) 45 { 46 if(x<=l && y>=r) 47 { 48 ma = max(ma,st_max[rt]); 49 mi = min(mi,st_min[rt]); 50 return ma - mi; 51 } 52 53 int mid = (l+r)>>1; 54 if(y<=mid) query(rt*2,l,mid,x,y,ma,mi); 55 else if(x>=mid+1) query(rt*2+1,mid+1,r,x,y,ma,mi); 56 else query(rt*2,l,mid,x,mid,ma,mi),query(rt*2+1,mid+1,r,mid+1,y,ma,mi); 57 58 return ma - mi; 59 } 60 61 int test(int len, int n, int k) 62 { 63 for(int i = len-1; i<n; i++) 64 { 65 LL ma = -LNF, mi = LNF; 66 if(query(1,0,n-1, i-len+1, i, ma,mi)<=1LL*k) 67 return 1; 68 } 69 return 0; 70 } 71 72 int main() 73 { 74 #ifdef LOCAL 75 freopen("input.txt", "r", stdin); 76 // freopen("output.txt", "w", stdout); 77 #endif // LOCAL 78 79 int n, k; 80 while(scanf("%d%d", &n, &k)!=EOF) 81 { 82 for(int i=0;i<n;i++) 83 scanf("%lld", &a[i]); 84 85 build(1,0,n-1); 86 87 int l = 1, r = n; 88 while(l<=r) 89 { 90 int mid = (l+r)/2; 91 if(test(mid, n,k)) 92 l = mid+1; 93 else 94 r = mid-1; 95 } 96 printf("%d ", r); 97 } 98 return 0; 99 }