Optimize!

這個題目代碼看了我半天。。

我們把終點關註在b數組， 我們先將b[ i ] 變成 h - b[ i ]並排好序， 對於一個a[ j ]來說如果它能和b[ i ]匹配， 那麽它能和b[ k ], k < i, 匹配。

什麽情況下能匹配成功呢， 就是b數組中 前 i 個數至少能和 len - i + 1, 個數匹配，這個隨便想想就知道， 或者通過霍爾定理也能很快得到，

那麽就能用線段樹取維護了。

#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 = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
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 n, len, h;
int a[N], b[N], c[N];

struct SegmentTree {
#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
    int mx[N << 2], lazy[N << 2];
    inline void pull(int rt) {
        mx[rt] = max(mx[rt << 1], mx[rt << 1 | 1]);
    }
    inline void push(int rt) {
        if(lazy[rt]) {
            mx[rt << 1] += lazy[rt];
            mx[rt << 1 | 1] += lazy[rt];
            lazy[rt << 1] += lazy[rt];
            lazy[rt << 1 | 1] += lazy[rt];
            lazy[rt] = 0;
        }
    }
    void build(int l, int r, int rt) {
        if(l == r) {
            mx[rt] = len - l + 1;
            return;
        }
        int mid = l + r >> 1;
        build(lson); build(rson);
        pull(rt);
    }
    void update(int L, int R, LL val, int l, int r, int rt) {
        if(R < l || r < L || R < L) return;
        if(L <= l && r <= R) {
            mx[rt] += val;
            lazy[rt] += val;
            return;
        }
        push(rt);
        int mid = l + r >> 1;
        update(L, R, val, lson);
        update(L, R, val, rson);
        pull(rt);
    }
} Tree;

int main() {
    scanf("%d%d%d", &n, &len, &h);
    for(int i = 1; i <= len; i++) {
        scanf("%d", &b[i]);
        b[i] = h - b[i];
    }
    sort(b + 1, b + 1 + len);
    for(int i = 1; i <= n; i++) {
        scanf("%d", &a[i]);
        c[i] = upper_bound(b + 1, b + 1 + len, a[i]) - b - 1;
    }
    Tree.build(1, len, 1);
    for(int i = 1; i <= len; i++) Tree.update(1, c[i], -1, 1, len, 1);
    int ans = 0;
    for(int i = 1; i + len - 1 <= n; i++) {
        ans += Tree.mx[1] <= 0;
        Tree.update(1, c[i], 1, 1, len, 1);
        if(i + len <= n) Tree.update(1, c[i + len], -1, 1, len, 1);
    }
    printf("%d\n", ans);
    return 0;
}

/*
*/

Codeforces 338E Optimize! 線段樹