题解

波兰人的j是苹果，p是橘子
还真是跟中国过不去啊= =写的时候很难受

我们先求出每个点作为起点，能延伸到的最大长度，这个可以处理成前缀和，查询一下区间最小值是不是小于0，用st表实现，如果区间最小值大于等于0，那么这段区间，以该点作为起点是合法的

然后求出每个点作为终点能延伸到的最大长度，处理成后缀和

然后枚举每个点作为起点，找出能延伸的最长右端点，然后再次二分答案，如果二分到的是mid，且mid到ri[i]中的最小值小于等于该点，那么右端点一定可以再往右，否则往左

三次二分答案，三次st表，比较神奇。。。

代码

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <cmath>
#include <bitset>
#include <queue>
#define enter putchar('\n')
#define space putchar(' ')
//#define ivorysi
#define pb push_back
#define mo 974711
#define pii pair<int,int>
#define mp make_pair
#define fi first
#define se second
#define MAXN 100005
#define eps 1e-12
using namespace std;
typedef long long int64;
typedef double db;
template<class T>
void read(T &res) {res = 0;char c = getchar();T f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {res = res * 10 - '0' + c;c = getchar();}res = res * f;
}
template<class T>
void out(T x) {if(x < 0) {x = -x;putchar('-');}if(x >= 10) out(x / 10);putchar('0' + x % 10);
}
int N,sum[MAXN],st[MAXN][20],len[MAXN],ri[MAXN],le[MAXN];
char s[MAXN];
int Qmin(int l,int r) {int t = len[r - l + 1];return min(st[l][t],st[r - (1 << t) + 1][t]);
}
void Solve() {read(N);scanf("%s",s + 1);for(int i = 1 ; i <= N ; ++i) {if(s[i] == 'j') sum[i] = sum[i - 1] - 1;else sum[i] = sum[i - 1] + 1;st[i][0] = sum[i];}for(int i = 2 ; i <= N ; ++i) len[i] = len[i / 2] + 1;for(int j = 1 ; j <= 18 ; ++j) {for(int i = 1 ; i <= N ; ++i) {if(i + (1 << j) - 1> N) break;st[i][j] = min(st[i][j - 1],st[i + (1 << j - 1)][j - 1]);}}for(int i = 1 ; i <= N ; ++i) {if(s[i] == 'j') continue;int L = i,R = N;while(L < R) {int mid = (L + R + 1) >> 1;if(Qmin(i,mid) - sum[i - 1] >= 0) L = mid;else R = mid - 1;}ri[i] = R;}memset(sum,0,sizeof(sum));for(int i = N ; i >= 1 ; --i) {if(s[i] == 'j') sum[i] = sum[i + 1] - 1;else sum[i] = sum[i + 1] + 1;st[i][0] = sum[i];}for(int j = 1 ; j <= 18 ; ++j) {for(int i = 1 ; i <= N ; ++i) {if(i + (1 << j) - 1 > N) break;st[i][j] = min(st[i][j - 1],st[i + (1 << j - 1)][j - 1]);}}for(int i = 1 ; i <= N ; ++i) {if(s[i] == 'j') {le[i] = N;continue;}int L = 1,R = i;while(L < R) {int mid = (L + R) >> 1;if(Qmin(mid,i) - sum[i + 1] >= 0) R = mid;else L = mid + 1;}le[i] = L;}for(int i = 1 ; i <= N ; ++i) st[i][0] = le[i];for(int j = 1 ; j <= 18 ; ++j) {for(int i = 1 ; i <= N ; ++i) {if(i + (1 << j) - 1 > N) break;st[i][j] = min(st[i][j - 1],st[i + (1 << j - 1)][j - 1]);}}int ans = 0;for(int i = 1 ; i <= N ; ++i) {if(ri[i] >= i) {int L = i,R = ri[i];while(L < R) {int mid = (L + R + 1) >> 1;if(Qmin(mid,ri[i]) <= i) L = mid;else R = mid - 1;}ans = max(ans,R - i + 1);}}out(ans);enter;
}
int main() {
#ifdef ivorysifreopen("f1.in","r",stdin);
#endifSolve();
}