Codeforces Round #278 (Div. 1) Strip RMQ

B. Strip

time limit per test 1 second

memory limit per test 256 megabytes

input standard input

output standard output

Alexandra has a paper strip with n numbers on it. Let's call them a_i from left to right.

Now Alexandra wants to split it into some pieces (possibly 1). For each piece of strip, it must satisfy:

Each piece should contain at least l numbers.
The difference between the maximal and the minimal number on the piece should be at most s.

Please help Alexandra to find the minimal number of pieces meeting the condition above.

Input

The first line contains three space-separated integers n, s, l (1 ≤ n ≤ 10⁵, 0 ≤ s ≤ 10⁹, 1 ≤ l ≤ 10⁵).

The second line contains n integers a_i separated by spaces ( - 10⁹ ≤ a_i ≤ 10⁹).

Output

Output the minimal number of strip pieces.

If there are no ways to split the strip, output -1.

Examples

Input

7 2 2
1 3 1 2 4 1 2

Output

Input

7 2 2
1 100 1 100 1 100 1

Output

-1

Note

For the first sample, we can split the strip into 3 pieces: [1, 3, 1], [2, 4], [1, 2].

For the second sample, we can't let 1 and 100 be on the same piece, so no solution exists.

题目大意：让你把一个序列分几段，每段的最大值和最小值之差不超过s，该段的长度也必须大于等于l，问你最少能分成几段？

这当题当然是要dp啦！反正我没想出来怎么划分。这题真是很棒!

首先给出dp公式 dp[i] = min{dp[j]+1, dp[i]}, i-j >= l && max{j+1~i}-min{j+1，~i}<= s

满足这个条件就可以dp了！用rmq求最大最小的差值，双指针维护左右两个点i，j序列。

#include <bits/stdc++.h>

using namespace std;
const int MAXN = 100005;
const int INF = 0x3f3f3f3f;
int f[MAXN][20], g[MAXN][20], a[MAXN], dp[MAXN];
int n, q;
void Rmq() {
    int lg = floor(log10(double(n))/log10(double(2)));
    for(int j = 1; j <= lg; ++j) {
        for(int i = 1; i <= n+1-(1<<j); ++i) {
            g[i][j] = max(g[i][j-1], g[i+(1<<(j-1))][j-1]);
            f[i][j] = min(f[i][j-1], f[i+(1<<(j-1))][j-1]);
        }
    }
}
int query(int a, int b) {

    int lg = floor(log10(double(b-a+1))/log10(double(2)));
    return max(g[a][lg], g[b-(1<<lg)+1][lg])
           -  min(f[a][lg], f[b-(1<<lg)+1][lg]);
}
int main() {
    int l, s;
    scanf("%d%d%d", &n, &s, &l);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        f[i][0] = g[i][0] = a[i];
        dp[i] = INF;
    }
    Rmq();
    dp[0] = 0;
    int pre = 0;
    for(int i = l; i <= n; i++) {
        while((i-pre >= l && query(pre+1, i) > s )|| dp[pre] == INF) pre++;
        printf("%d %d\n", pre, i);
        if(i - pre >= l)
            dp[i] = min(dp[pre]+1, dp[i]);
    }
    if(dp[n] == INF) printf("-1\n");
    else printf("%d\n", dp[n]);
    return 0;
}

View Code

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