思路
- 因为涉及到 l 0 l_0 l0 最大连续长度和 l 1 l_1 l1 最大连续长度,考虑枚举 l 1 l_1 l1 的长度 d p l 0 dp\;l_0 dpl0 的长度。
- m a x 0 p r e f [ i ] [ c n t ] max0pref[i][cnt] max0pref[i][cnt] 表示以 i − 1 i-1 i−1 结尾的 l 0 l_0 l0
- m a x 0 s u f [ i ] [ c n t ] max0suf[i][cnt] max0suf[i][cnt] 表示以 i i i 开始的 l 0 l_0 l0
- m a x 0 b y 1 [ i ] max0by1[i] max0by1[i] 表示 l 1 l_1 l1 为 i i i 时最大的 l 0 l_0 l0
计算
- m a x 0 p r e f , m a x 0 s u f max0pref,max0suf max0pref,max0suf 都是 1 → 0 1\rightarrow0 1→0 ,所以记录 c n t cnt cnt 更改次数
for(int i = 0; i < n; ++i) {int cnt = 0;for(int j = i; j <= n; ++j) {cnt += s[j - 1] == '1';max0pref[j][cnt] = max(max0pref[j][cnt], j - i);max0suf[i][cnt] = max(max0suf[i][cnt], j - i);}
}
更新
- m a x 0 p r e f i , j = { m a x 0 p r e f i − 1 , j m a x 0 p r e f i , j − 1 max0pref_{i,j}=\begin{cases}max0pref_{i-1,j}\\max0pref_{i,j-1}\end{cases} max0prefi,j={max0prefi−1,jmax0prefi,j−1
- m a x 0 s u f i , j = { m a x 0 s u f i + 1 , j m a x 0 s u f i , j − 1 max0suf_{i,j}=\begin{cases}max0suf_{i+1,j}\\max0suf_{i,j-1}\end{cases} max0sufi,j={max0sufi+1,jmax0sufi,j−1
for(int i = 0; i <= n; ++i) {for(int cnt = 0; cnt <= n; ++cnt) {if (i) max0pref[i][cnt] = max(max0pref[i][cnt], max0pref[i - 1][cnt]);if (cnt) max0pref[i][cnt] = max(max0pref[i][cnt], max0pref[i][cnt - 1]);}
}
for(int i = n; i >= 0; --i) {for(int cnt = 0; cnt <= n; ++cnt) {if (i + 1 <= n) max0suf[i][cnt] = max(max0suf[i][cnt], max0suf[i + 1][cnt]);if (cnt) max0suf[i][cnt] = max(max0suf[i][cnt], max0suf[i][cnt - 1]);}
}
更新 m a x 0 b y 1 max0by1 max0by1
- 每次选取一个区间,求当这个区间全为 1 1 1 时的 l 0 l_0 l0。
for(int i = 0; i < n; ++i) {int cnt=0;for(int j = i; j <= n; ++j) {if(j > i) cnt += s[j - 1] == '0';if(cnt > k) break;max0by1[j - i] = max(max0by1[j - i], max0pref[i][k - cnt]);max0by1[j - i] = max(max0by1[j - i], max0suf[j][k - cnt]);}
}
答案
- 按思路枚举 l 1 l_1 l1 长度即可
for(int i = 0; i <= n; ++i) {for(int a = 1; a <= n; ++a) ans[a] = max(ans[a], i + max0by1[i] * a);
}
Thick Twice, Code Once
#include<bits/stdc++.h>
#define il inline
#define get getchar
#define put putchar
#define is isdigit
#define int long long
#define dfor(i,a,b) for(int i=a;i<=b;++i)
#define dforr(i,a,b) for(int i=a;i>=b;--i)
#define dforn(i,a,b) for(int i=a;i<=b;++i,put(10))
#define mem(a,b) memset(a,b,sizeof a)
#define memc(a,b) memcpy(a,b,sizeof a)
#define pr 114514191981
#define gg(a) cout<<a,put(32)
#define INF 0x7fffffff
#define tt(x) cout<<x<<'\n'
#define ls i<<1
#define rs i<<1|1
#define la(r) tr[r].ch[0]
#define ra(r) tr[r].ch[1]
#define lowbit(x) (x&-x)
using namespace std;
typedef unsigned int ull;
typedef pair<int ,int > pii;
int read(void)
{int x=0,f=1;char c=get();while(!is(c)) (f=c==45?-1:1),c=get();while(is(c)) x=(x<<1)+(x<<3)+(c^48),c=get();return x*f;
}
void write(int x)
{if(x<0) x=-x,put(45);if(x>9) write(x/10);put((x%10)^48);
}
#define writeln(a) write(a),put(10)
#define writesp(a) write(a),put(32)
#define writessp(a) put(32),write(a)
const int N=1e6+10,M=3e4+10,SN=5e3+10,mod=998244353;
int n, k;
char s[N];
signed main()
{int T = read();while(T--){n = read(), k = read();scanf("%s", s);vector<int > max0by1(n + 1,-INF);vector< vector<int > > max0pref(n + 1, vector<int >(n + 1));vector< vector<int > > max0suf(n + 1, vector<int >(n + 1));for(int i = 0; i < n; ++i) {int cnt = 0;for(int j = i + 1; j <= n; ++j) {cnt += s[j - 1] == '1';max0pref[j][cnt] = max(max0pref[j][cnt], j - i);max0suf[i][cnt] = max(max0suf[i][cnt], j - i);}}for(int i = 0; i <= n; ++i) {for(int cnt = 0; cnt <= n; ++cnt) {if (i) max0pref[i][cnt] = max(max0pref[i][cnt], max0pref[i - 1][cnt]);if (cnt) max0pref[i][cnt] = max(max0pref[i][cnt], max0pref[i][cnt - 1]);}}for(int i = n; i >= 0; --i) {for(int cnt = 0; cnt <= n; ++cnt) {if (i + 1 <= n) max0suf[i][cnt] = max(max0suf[i][cnt], max0suf[i + 1][cnt]);if (cnt) max0suf[i][cnt] = max(max0suf[i][cnt], max0suf[i][cnt - 1]);}}vector<int > ans(n + 1, -INF);for(int i = 0; i < n; ++i) {int cnt=0;for(int j = i; j <= n; ++j) {if(j > i) cnt += s[j - 1] == '0';if(cnt > k) break;max0by1[j - i] = max(max0by1[j - i], max0pref[i][k - cnt]);max0by1[j - i] = max(max0by1[j - i], max0suf[j][k - cnt]);}}for(int i = 0; i <= n; ++i) {for(int a = 1; a <= n; ++a) ans[a] = max(ans[a], i + max0by1[i] * a);}for(int i = 1; i <= n; ++i) writesp(ans[i]);puts("");}return 0;
}