Codeforces Round #278 (Div. 1) B

B - Strip

思路：简单dp，用st表+单调队列维护一下。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define PII pair<int, int>
#define y1 skldjfskldjg
#define y2 skldfjsklejg
using namespace std;

const int N = 1e5 + 7;
const int M = 1e7 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1000000007;

int n, s, l, tot, head, rear, a[N], dp[N], stk[N], Log[N];

struct ST {
    int dp[N][20],ty;
    void build(int n, int b[], int _ty) {
        ty = _ty;
        for(int i = 1; i <= n; i++) dp[i][0] = ty * b[i];
        for(int j = 1; j <= Log[n]; j++)
            for(int i = 1; i+(1<<j)-1 <= n; i++)
                dp[i][j] = max(dp[i][j-1], dp[i+(1<<(j-1))][j-1]);
    }
    int query(int x, int y) {
        int k = Log[y - x + 1];
        return ty * max(dp[x][k], dp[y-(1<<k)+1][k]);
    }
}mx, mn;

bool check(int L, int R) {
    return mx.query(L, R) - mn.query(L, R) <= s;
}

int main() {

    for(int i = -(Log[0]=-1); i < N; i++)
        Log[i] = Log[i - 1] + ((i & (i - 1)) == 0);

    memset(dp, -1, sizeof(dp));

    scanf("%d%d%d", &n, &s, &l);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
    mx.build(n, a, 1); mn.build(n, a, -1);

    head = 1, rear = 0;
    for(int i = 1; i <= n; i++) {
        if(i >= l && check(1, i)) {
            dp[i] = 1;
            stk[++rear] = i;
            continue;
        }
        while(rear >= head && !check(stk[head] + 1, i)) head++;
        if(rear >= head && i - stk[head] >= l) {
            dp[i] = dp[stk[head]] + 1;
            stk[++rear] = i;
        }
    }
    printf("%d
", dp[n]);
    return 0;
}

/*
76 63 14
89 87 35
20 15 56
*/