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Hotaru‘s problem
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1765 Accepted Submission(s): 635
Problem Description
Hotaru Ichijou recently is addicated to math problems. Now she is playing with N-sequence.
Let‘s define N-sequence, which is composed with three parts and satisfied with the following condition:
1. the first part is the same as the thrid part,
2. the first part and the second part are symmetrical.
for example, the sequence 2,3,4,4,3,2,2,3,4 is a N-sequence, which the first part 2,3,4 is the same as the thrid part 2,3,4, the first part 2,3,4 and the second part 4,3,2 are symmetrical.
Give you n positive intergers, your task is to find the largest continuous sub-sequence, which is N-sequence.
Input
There are multiple test cases. The first line of input contains an integer T(T<=20), indicating the number of test cases.
For each test case:
the first line of input contains a positive integer N(1<=N<=100000), the length of a given sequence
the second line includes N non-negative integers ,each interger is no larger than 109 , descripting a sequence.
Output
Each case contains only one line. Each line should start with “Case #i: ”,with i implying the case number, followed by a integer, the largest length of N-sequence.
We guarantee that the sum of all answers is less than 800000.
Sample Input
Sample Output
题目大意:给你t组数据,一个n,下面一行有n个数,问你能形成形如abccbaabc这样的序列长度最长是多少。
解题思路:先用manacher处理出来串的所有字符的回文半径。然后枚举第一段跟第二段回文位置i,从i+p[i]-->i进行枚举第二段跟第三段回文位置j。如果以j为回文中心的左端能小于等于i的位置,说明满足要求,更新结果。
#include<bits/stdc++.h>
using namespace std;
#define min(a,b) ((a)<(b)?(a):(b))
const int maxn=1e6;
int a[maxn],p[maxn];
void Manacher(int n){
a[0]=-2;a[n+1]=-1;a[n+2]=-3;
int mx=0,id=0;
for(int i=1;i<=n+1;i++){ //需要处理到n+1
if(i<mx){
p[i]=min(p[id*2-i],mx-i); //这里写的时候写成mx-id,SB了。
}else{
p[i]=1;
}
for(;a[i+p[i]]==a[i-p[i]];++p[i]); //-2,-3防越界
if(i+p[i]>mx){
mx=p[i]+i;
id=i;
}
}
for(int i=1;i<=n+1;++i){
--p[i];
}
}
int main(){
// freopen("1003.in","r",stdin);
// freopen("OUTTTT.txt","w",stdout);
int t,n,cnt=0;
scanf("%d",&t);
while(t--){
scanf("%d",&n);
for(int i=1;i<=2*n;i+=2){
a[i]=-1;
scanf("%d",&a[i+1]);
}
Manacher(2*n);
int maxv=0,j;
for(int i=1;i<=2*n;i+=2){ //2*n
for(j=i+p[i];j-maxv>i;j-=2){ //逆序枚举。manacher算法保证j<=2*n+1
if(j-p[j]<=i){
maxv=j-i;
break;
}
}
}
maxv=3*(maxv/2);
printf("Case #%d: %d\n",++cnt,maxv);
}
return 0;
}
HDU 5371——Hotaru's problem——————【manacher处理回文】
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原文地址:http://www.cnblogs.com/chengsheng/p/4726693.html