暴力 DP 做这题只有 30 分
考虑用线段树优化这个 DP
先处理一下整个房间都膜拜一个人的情况, 然后将 1 的当成 -1, 2 当成 1, 处理前缀和, 可以发现对于前缀和为 x 的情况, 只能从前缀和为 [x - k, x + k] 的地方转移过来, 用线段树维护 DP 数组的最小值就行了
- #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;
- }
来源: http://www.bubuko.com/infodetail-2759350.html