CF1204D Kirk and a Binary String

D. Kirk and a Binary String

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題意

給出一個01串S，長度$\leq 100000$

請你構建一個01串P，長度與S相同

使得對於任意l,r，$1\leq l \leq r \leq S$

對於區間[l,r]的最長不下降子串長度相同

思路

• 由於只有1變爲0，若[l,r]改變，[1,n]必定改變
  • 我們可以將S分成若干的01串，且每個串前面都爲0，後面都爲1
  • 比如將0111001101分成0111，0011，01
  • [1,n]的LIS即爲將每個段中取出0或1
  • [L,R]的LIS求法與[1,n]同，若[L,R]改變，則必定有01分串改變
• 取上面的逆否命題，若[1,n]不變，[l,r]不變

對於區間[l,r]，若將位置i的1改爲0

新的最長不下降子序列長度爲：

設LIS(l,r)爲l到r的最長不下降子序列，sum[x][l,r]爲l到r中x的個數

• 若[l,i]均爲0，顯然可以改變爲0，無影響

• $LIS(l,r)=max(LIS(l,i-1)+sum[1][i+1][r],sum[0][i-1]+1+LIS(i+1,r))$

  前面的LIS加上後面的1或者後面的LIS加上前面的0

• 顯然改變後，只改變了0,1個數，LIS[l][r]不改變

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代碼

預處理LIS

int cnt = 0;
for (int i = 1; i <= n; pre_dp[i] = cnt, i++) {
	if (d[cnt] <= res[i]) d[++cnt] = res[i];
	else d[upper_bound(d + 1, d + 1 + cnt, res[i]) - d] = res[i];
}
cnt = 0;
d[cnt] = 2;
for (int i = n; i >= 1; suf_dp[i] = cnt, i--) {
	if (d[cnt] >= res[i]) d[++cnt] = res[i];
	else d[upper_bound(d + 1, d + 1 + cnt, res[i],greater<int>()) - d] = res[i];
}

solve

int pre_zero = 0, suf_one = 0;
for (int i = 2; i <= n; suf_one += res[i] == 1, i++);
int lis = pre_dp[n];
for (int i = 1; i <= n; i++) {
	if (res[i] && lis == max(pre_dp[i - 1] + suf_one,
	                         pre_zero + 1 + suf_dp[i + 1])) {
		res[i] = 0;
		pre_zero++;
		suf_one--;
	} else {
		pre_zero += res[i] == 0;
		suf_one -= res[i] == 1;
	}
}

AC

char s[maxn];
int res[maxn];
int pre_dp[maxn], suf_dp[maxn];
int d[maxn];
int main() {
	scanf("%s", s + 1);
	int n = strlen(s + 1);
	for (int i = 1; i <= n; i++) if (s[i] == '1')res[i] = 1;
	int cnt = 0;
	for (int i = 1; i <= n; pre_dp[i] = cnt, i++) {
		if (d[cnt] <= res[i]) d[++cnt] = res[i];
		else d[upper_bound(d + 1, d + 1 + cnt, res[i]) - d] = res[i];
	}
	cnt = 0; d[cnt] = 2;
	for (int i = n; i >= 1; suf_dp[i] = cnt, i--) {
		if (d[cnt] >= res[i]) d[++cnt] = res[i];
		else d[upper_bound(d + 1, d + 1 + cnt, res[i],greater<int>()) - d] = res[i];
	}
	int pre_zero = 0, suf_one = 0; for (int i = 2; i <= n; suf_one += res[i] == 1, i++);
	int lis = pre_dp[n];
	for (int i = 1; i <= n; i++) {
		if (res[i] && lis == max(pre_dp[i - 1] + suf_one,
			pre_zero + 1 + suf_dp[i + 1])) {
			res[i] = 0; pre_zero++; suf_one--;
		}
		else {
			pre_zero += res[i] == 0;
			suf_one -= res[i] == 1;
		}
	}
	for (int i = 1; i <= n; i++) s[i] = res[i] + '0';
	printf("%s\n", s + 1);
}