bzoj4236 JOIOJI

题目

显然转化为前缀再看相等即可。

sj[j]-sj[i]==so[j]-so[i]==si[j]-si[i]

转化一下

(sj[j]-so[j])-(sj[j]-so[i])==(si[j]-so[j])-(si[i]-so[i])

这样，就转化成两维的了，排一下序就可以统计出来了。

#include<bits/stdc++.h>#define N 200005using namespace std;int n,sj[N+1],so[N+1],si[N+1],ans,last;struct data{    int x;    int y;    int id;};data A[N+1];char x;inline char nc(){    static char buf[100000],*p1=buf,*p2=buf;    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;}inline int read(){    int x=0,b=1;    char c=nc();    for(;!(c<='9'&&c>='0');c=nc())if(c=='-')b=-1;    for(;c<='9'&&c>='0';c=nc())x=x*10+c-'0';    return x*b;}inline char get(){    char c=nc();    for(;!(c<='Z'&&c>='A');)c=nc();    return c;}inline bool cmp(const data &A,const data &B){    if(A.x!=B.x)return A.x<B.x;    if(A.y!=B.y)return A.y<B.y;    return A.id<B.id;}int main(){    //freopen("in.txt","r",stdin);    n=read();    for(int i=1;i<=n;i++)    {        x=get();        sj[i]=sj[i-1]+(x=='J');        si[i]=si[i-1]+(x=='I');        so[i]=so[i-1]+(x=='O');    }    for(int i=0;i<=n;i++)    {        A[i].x=si[i]-so[i];        A[i].y=sj[i]-so[i];        A[i].id=i;    }    sort(A,A+n+1,cmp);    ans=0,last=0;    for(int i=1;i<=n;i++)        if(A[i].x==A[i-1].x&&A[i].y==A[i-1].y)ans=max(A[i].id-A[last].id,ans);        else last=i;    cout<<ans;    return 0;}