zoj 3349 dp + 线段树优化
题目：给出一个序列，找出一个最长的子序列，相邻的两个数的差在d以内。
1 /* 2 线段树优化dp 3 dp[i]表示前i个数的最长为多少，则dp[i]=max(dp[j]+1) abs(a[i]-a[j])<=d 4 复杂度为O(n ^ 2) 5 利用线段树优化，线段树保存区间最大值。离散化后便可求出，还要注意 对于叶子节点保存的即为dp的值，每次更改即可，开始一直累加。。。。。 6 */ 7 #include <iostream> 8 #include <cstdio> 9 #include <cstring> 10 #include <cmath> 11 #include <algorithm> 12 13 using namespace std; 14 #define lson l,m,rt<<1 15 #define rson m + 1, r, rt<<1|1 16 const int maxn = 1e5 + 5; 17 int s[maxn]; 18 int n, d; 19 int san[maxn], tot; 20 int sum[maxn << 2]; 21 void pushUp(int rt){ 22 sum[rt] = max(sum[rt<<1], sum[rt<<1|1]); 23 } 24 void update(int pos, int c, int l, int r, int rt){ 25 if (l == r){ 26 sum[rt] = c;//注意 27 return ; 28 } 29 int m = (l + r) >> 1; 30 if (pos <= m) update(pos, c, lson); 31 else update(pos, c, rson); 32 pushUp(rt); 33 } 34 int query(int L, int R, int l, int r, int rt){ 35 if (L <= l && R >= r){ 36 return sum[rt]; 37 } 38 int m = (l + r) >> 1; 39 int ret = 0; 40 if (L <= m) ret = query(L, R, lson); 41 if (R > m) ret = max(ret, query(L, R, rson)); 42 return ret; 43 } 44 int main(){ 45 while (~scanf("%d%d", &n, &d)){ 46 tot = 0; 47 for (int i = 1; i <= n; ++i){ 48 scanf("%d", &s[i]); 49 san[tot++] = s[i]; 50 } 51 sort(san, san + tot); 52 tot = unique(san, san + tot) - san; 53 54 memset(sum, 0, sizeof(sum)); 55 int ans = 0; 56 for (int i = 1; i <= n; ++i){ 57 int pos = lower_bound(san, san + tot, s[i]) - san + 1; 58 int l = lower_bound(san, san + tot, s[i] - d) - san + 1; 59 int r = upper_bound(san, san + tot, s[i] + d) - san; 60 int que = query(l, r, 1, tot, 1) + 1; 61 //cout << " l = " << l << " r = " << r << endl; 62 ans = max(ans, que); 63 //cout << " ans = " << ans << endl; 64 update(pos, que, 1, tot, 1); 65 } 66 printf("%d\n", ans); 67 } 68 return 0; 69 }
