题解 | #L-最长连续相同字符#

L-最长连续相同字符,分段统计

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <vector>
#include <iostream>
using namespace std;
int a[400][400];
int b[400], c[400], a1[400];
int n,n1,last,ans,ansi,ans1=0,ansi1=0;
char str[101000];
void count(int x){
    int j,i=x*n1,y=min(i+n1-1,n-1);
    a1[x] = 0;b[x] = 0;c[x] =0;
    while(i<=y){
        j=i;
        while(j<=y && str[j] ==str[i]) ++j;
        a[x][a1[x]] =j-i;
        ++a1[x];
        if (j-i>b[x]){
            b[x] = j-i;c[x] = i;
        }
        i=j;
    }
    return;
}
void count1(int l, int r){
    int i=l,j;
    while(i<=r){
        j=i;
        while(j<=r && str[j] ==str[i]) ++j;
        last = j-i;
        if (last>ans){
            ans = last;ansi=i;
        }
        i=j;
    }
}
int count2(int l, int r){
    int i=r,j, last1;
    last1 = 0;ans1=0,ansi1=0;
    while(i>=l){
        j=i;
        while(j>=l && str[j] ==str[i]) --j;
        last1 = i-j;
        if (last1>=ans1){
            ans1 = last1;ansi1=j+1;
        }
        i=j;
    }
    return last1;
}
void my_ans(){
    int i,j,m,k,op,l,r,mi,ma,mid,x,y,last1;
    cin>>n>>m;
    char ch;
    n1 = (int)sqrt(n)+1;
    memset(b,0,sizeof(b));
    memset(c,0,sizeof(c));
    for(i=0;i<n;++i) cin>>str[i];
    for(i=0;i<n1;++i) count(i);
    while(m>0){
        --m;cin>>op;
        if (op==1){
            cin>>l>>r;--l;--r;
            ans =1;ansi=l;last=0;
            if (r-l<8000){
                count1(l, r);
            }else{
                count1(l, (l/n1)*n1+n1-1);
                i=l/n1+1;
                while(i<r/n1){
                    if (str[i*n1] == str[i*n1-1]){
                        if (last+a[i][0]>ans){
                            ans = last+a[i][0];ansi = i*n1-last;
                        }
                        last += a[i][0];
                    }
                    if (a1[i] >1) last = a[i][a1[i]-1];
                    if (b[i]>ans){
                        ans = b[i];ansi=c[i];
                    }
                    ++i;
                }
                last1 = count2((r/n1)* n1, r);
                if (str[(r/n1)* n1] == str[(r/n1)* n1-1]){
                    if (last1+last>ans){
                        ans = last1+last;ansi = (r/n1)* n1 -last;
                    }
                }
                if (ans1>ans){
                    ans = ans1;ansi=ansi1;
                }
            }
            cout<<ansi+1<<" "<<ansi+ans<<endl;
        }else{
            cin>>x>>ch;--x;
            str[x] = ch;
            count(x/n1);
        }
    }
    return;
}
int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int t=1,i,j;
    //cin>>t;
    while(t>0){
        --t;my_ans();
    }
    return 0;
}
// 64 位输出请用 printf("%lld")