https://www.luogu.org/problemnew/show/P2418

暴力 DP 做這題只有 30 分

考慮用線段樹優化這個 DP

先處理一下整個房間都膜拜一個人的情況，然後將 1 的當成 -1， 2 當成 1，處理前綴和，可以發現對於前綴和為 x 的情況，只能從前綴和為 [x - k, x + k] 的地方轉移過來，用線段樹維護 DP 數組的最小值就行了

#include <bits/stdc++.h>
using namespace std;

const int N = 500000 + 10;

int minn[N << 3], a[N], s[N], f[N];
int n, k, last;

void change(int u, int l, int r, int x, int y) {
    minn[u] = min(minn[u], y);
    if(l == r) return;
    int mid = (l + r) >> 1;
    if(mid >= x) change(u << 1, l, mid, x, y);
    else change(u << 1 | 1, mid + 1, r, x, y); 
}

int Q;

void query(int u, int l, int r, int L, int R) {
    if(l <= L && R <= r) {
        Q = min(Q, minn[u]);
        return;
    }
    int mid = (L + R) >> 1;
    if(mid >= l) query(u << 1, l, r, L, mid);
    if(mid + 1 <= r) query(u << 1 | 1, l, r, mid + 1, R);
}

int main() {
    memset(minn, 0x3f, sizeof(minn));
    cin >> n >> k; s[0] = N;
    for(int i = 1; i <= n; i++) {
        int t; scanf("%d", &t);
        if(t == 1) a[i] = -1;
        else a[i] = 1;
        s[i] = s[i - 1] + a[i]; 
    }
    change(1, 1, n + k + N, N, 0);
    for(int i = 1; i <= n; i++) {
        if(a[i] != a[i - 1]) last = f[i - 1];
        else last = min(last, f[i - 1]);
        f[i] = last + 1;
        Q = INT_MAX;
        query(1, s[i] - k, s[i] + k, 1, n + k + N);
        f[i] = min(f[i], Q + 1);
        change(1, 1, n + k + N, s[i], f[i]);
    }
    cout << f[n] << endl;
    return 0;
}
 

luoguP2418 yyy loves OI IV