hdu 4513 吉哥系列故事——完美队形II (回文串,manacher)
题目链接：hdu4513
题意：给出一个n的数的序列，求出一个最长的回文字串，并且满足从[l,mid]单调增（非严格单调，可以相等），[mid,r]单调减（同样是可以相等）
思路：manacher...int型的也是可以搞的。。要求单调的话。。。while扩展的时候判一下就好了。。。
/* ***********************************************
Author :111qqz
Created Time :2016年04月18日 星期一 20时32分45秒
File Name :code/hdu/4513.cpp
************************************************ */
1#include <cstdio>
2#include <cstring>
3#include <iostream>
4#include <algorithm>
5#include <vector>
6#include <queue>
7#include <set>
8#include <map>
9#include <string>
10#include <cmath>
11#include <cstdlib>
12#include <ctime>
13#define fst first
14#define sec second
15#define lson l,m,rt<<1
16#define rson m+1,r,rt<<1|1
17#define ms(a,x) memset(a,x,sizeof(a))
18typedef long long LL;
19#define pi pair < int ,int >
20#define MP make_pair
1using namespace std;
2const double eps = 1E-8;
3const int dx4[4]={1,0,0,-1};
4const int dy4[4]={0,-1,1,0};
5const int inf = 0x3f3f3f3f;
6const int N=2E5+7;
7int n;
8int p[N];
9int a[N];
1int manacher( int *s)
2{
3 for ( int i = n ; i >= 0 ; i--)
4 {
5 s[i+i+2] = s[i];
6 s[i+i+1] = 300;
7 }
8 s[0] = 400;s[2*n+3] =405;
1 // for ( int i = 1 ; i < 2*n+1 ; i++) cout<<"s[i]:"<<s[i]<<endl;
2 int id = 0 ;
3 int maxlen = 0 ;
1 for ( int i = 2 ; i < 2*n +1 ; i++)
2 {
3 if (p[id]+id>i) p[i] = min(p[2*id-i],p[id]+id-i);
4 else p[i] = 1;
1 while (s[i-p[i]]==s[i+p[i]])
2 {
3 if (s[i-p[i]]==300)
4 {
5 p[i]++;
6 }
7 else if (s[i-p[i]]<300)
8 {
9 if (s[i-p[i]]>s[i-p[i]+2]) break;
10 p[i]++;
11 }
1// cout<<"aaaaaaaaaooooooo"<<endl;
2 }
3// cout<<"p[i]:"<<p[i]<<endl;
4 if (id+p[id]<i+p[i]) id = i ;
if (maxlen<p[i]) maxlen = p[i];
}
1 return maxlen-1;
2}
3int main()
4{
5 #ifndef ONLINE_JUDGE
6 freopen("code/in.txt","r",stdin);
7 #endif
8 int T;
9 scanf("%d",&T);
10 while (T--)
11 {
12 scanf("%d",&n);
13 for ( int i = 0 ; i < n ; i++) scanf("%d",&a[i]);
1 int ans =manacher(a);
2 printf("%d\n",ans);
3 }
1 #ifndef ONLINE_JUDGE
2 fclose(stdin);
3 #endif
4 return 0;
5}