Codeforces 875E Delivery Club dp

Delivery Club

首先二分答案， 我们考虑dp去check

我们可以发现所有状态要从两个点在 i 和 i - 1转移过来。

所以我们令dp[ i ] 表示 能否到达两个快递员在分别 i 号点 和 i - 1号点的状态。

转移就是把在 i 号位置的快递员不断地往后移， 如果能移到 j && abs(a[ j  + 1 ] - a[ j ]) <= d   则 dp[ j + 1 ] = true;

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}

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

int n, s1, s2;
int a[N];
int dp[N];
ST rmqmx, rmqmn;

bool check(int d) {
    memset(dp, 0, sizeof(dp));
    dp[2]++;
    dp[3]--;
    for(int i = 2; i <= n; i++) {
        dp[i] += dp[i - 1];
        if(!dp[i] || abs(a[i] - a[i - 1]) > d) continue;
        if(i == n) return true;

        int p1 = i, p2 = i;

        int low = i, high = n, mid;
        while(low <= high) {
            mid = low + high >> 1;
            if(rmqmx.query(i, mid) <= a[i - 1] + d) p1 = mid, low = mid + 1;
            else high = mid - 1;
        }
        low = i, high = n;
        while(low <= high) {
            mid = low + high >> 1;
            if(rmqmn.query(i, mid) >= a[i - 1] - d) p2 = mid, low = mid + 1;
            else high = mid - 1;
        }
        int p = min(p1, p2);
        if(p == n) return true;
        dp[i + 1]++; dp[p + 2]--;
    }
    return false;
}

int main() {
    scanf("%d%d%d", &n, &s1, &s2);
    n += 2;
    for(int i = 3; i <= n; i++) scanf("%d", &a[i]);
    a[1] = s1; a[2] = s2;
    rmqmx.build(n, a, 1);
    rmqmn.build(n, a, -1);
    int low = abs(s1 - s2), high = 1000000000, ans = -1;
    while(low <= high) {
        int mid = low + high >> 1;
        if(check(mid)) ans = mid, high = mid - 1;
        else low = mid + 1;
    }
    printf("%d\n", ans);
    return 0;
}

/*
*/