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HDU 4513 吉哥系列故事 完美队形II (manacher)

2015-11-03 22:16 453 查看
题意:

N<=105的串,求最长回文串,并且左边的部分是不降的

分析:

计算p数组的时候判断一下就好了,−−注意′#′

代码:

//
//  Created by TaoSama on 2015-11-03
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 2e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, k, p
;
int a
, s
;

int manacher() {
s[0] = '@'; s[1] = '#'; n = 2;
for(int i = 0; i < k; ++i)
s[n++] = a[i], s[n++] = '#';
s
= 0;

int mx = 0, id, ret = 0;
for(int i = 1; i < n; ++i) {
p[i] = mx > i ? min(mx - i, p[2 * id - i]) : 1;
while(s[i - p[i]] == s[i + p[i]]) {
if(p[i] == 1 || s[i + p[i]] == '#') {}
else {
if(s[i + p[i]] > s[i + p[i] - 2]) break;
}
++p[i];
}
if(mx < i + p[i]) mx = i + p[i], id = i;
ret = max(ret, p[i] - 1);
}
return ret;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
while(t--) {
scanf("%d", &k);
for(int i = 0; i < k; ++i) scanf("%d", a + i);
printf("%d\n", manacher());
}
return 0;
}
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标签:  manacher