ZOJ3349——Special Subsequence

Special Subsequence

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Time Limit: 5 Seconds      Memory Limit: 32768 KB

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There a sequence S with n integers , and A is a special subsequence thatsatisfies |A_i-A_i-1| <=d ( 0 <i<=|A|))

Now your task is to find the longest special subsequence of a certain sequenceS

Input

There are no more than 15 cases , process till the end-of-file

The first line of each case contains two integer n and d ( 1<=n<=100000 ,0<=d<=100000000) as in the description.

The second line contains exact n integers , which consist the sequneceS .Eachinteger is in the range [0,100000000] .There is blank between each integer.

There is a blank line between two cases

Output

For each case , print the maximum length of special subsequence you can get.

Sample Input

5 21 4 3 6 55 01 2 3 4 5

Sample Output

31

水dp+水线段树

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#include <map>#include <set>#include <list>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int N = 100010;int dp[N], xis[N], arr[N];int n, d, cnt;struct node { int l, r; int val; }tree[N << 2]; int BinSearch(int x) { int l = 1, r = cnt, mid; while (l <= r) { mid = (l + r) >> 1; if (xis[mid] > x) { r = mid - 1; } else if (xis[mid] < x) { l = mid + 1; } else { break; } } return mid; } int left_binsearch(int x){int l = 1, r = cnt, mid, ans = -1;while (l <= r){mid = (l + r) >>1;if (xis[mid] >= x){ans = mid;r = mid - 1;}else{l = mid + 1;}}return ans;}int right_binsearch(int x){int l = 1, r = cnt, mid, ans = -1;while (l <= r){mid = (l + r) >>1;if (xis[mid] <= x){ans = mid;l = mid + 1;}else{r = mid - 1;}}return ans;} void build(int p, int l, int r) { tree[p].l = l; tree[p].r = r; tree[p].val = -1; if (l == r) { return; } int mid = (l + r) >> 1; build(p << 1, l, mid); build(p << 1 | 1, mid + 1, r); } void update(int p, int pos, int val) { if (tree[p].l == tree[p].r) { tree[p].val = max(tree[p].val, val); return; } int mid = (tree[p].l + tree[p].r) >> 1; if (pos <= mid) { update(p << 1, pos, val); } else { update(p << 1 | 1, pos, val); } tree[p].val = max(tree[p << 1].val, tree[p << 1 | 1].val); } int query(int p, int l, int r) { if (l <= tree[p].l && r >= tree[p].r) { return tree[p].val; } int mid = (tree[p].l + tree[p].r) >> 1; if (r <= mid) { return query(p << 1, l, r); } else if (l > mid) { return query(p << 1 | 1, l, r); } else { return max(query(p << 1, l, mid), query(p << 1 | 1, mid + 1, r)); } }int main(){while(~scanf("%d%d", &n, &d)){cnt = 0;int ans = 0;memset (dp, 0, sizeof(dp));for (int i = 1; i <= n; ++i){scanf("%d", &arr[i]);xis[++cnt] = arr[i];}sort (xis + 1, xis + cnt + 1);cnt = unique(xis + 1, xis + cnt + 1) - xis - 1;build(1, 1, cnt);for (int i = 1; i <= n; ++i){int cur = BinSearch(arr[i]);int l = left_binsearch(arr[i] - d);int r = right_binsearch(arr[i] + d);if (l < 0){l = 1;}if (r < 0){r = cnt;}int tmp = query(1, l, r);if (tmp == -1){dp[i] = 1;}else{dp[i] = tmp + 1;}update(1, cur, dp[i]);ans = max(ans, dp[i]);}printf("%d\n", ans);}return 0;}