HDU 3068 最长回文

HDU_3068

    因为做中欧区域赛的题目时，题解有提到Manacher's ALGORITHM，于是就学了一下并找了两个题练练手，一个是HDU_3068还有一个是URAL_1297。推荐一篇讲这个算法讲的感觉挺清楚的博客：http://www.felix021.com/blog/read.php?2040。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXD 220010
int N, p[MAXD];
char str[MAXD], b[MAXD];
void init()
{
    int i;
    for(i = 0; str[i]; i ++) b[2 * i + 1] = '#', b[2 * i + 2] = str[i];
    N = 2 * i + 1;
    b[0] = '$', b[N] = b[N + 1] = '#';
}
void solve()
{
    int i, id, max = 0, ans = 0;
    for(i = 1; i <= N; i ++)
    {
        p[i] = i < max ? std::min(max - i, p[2 * id - i]) : 1;    
        while(b[i + p[i]] == b[i - p[i]]) ++ p[i];
        if(i + p[i] > max) max = i + p[i], id = i;
        ans = std::max(ans, p[i] - 1);
    }
    printf("%d\n", ans);
}
int main()
{
    while(scanf("%s", str) == 1)
    {
        init();
        solve();    
    }
    return 0;    
}

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXD 220010
int N, p[MAXD];
char str[MAXD], b[MAXD];
void init()
{
    int i;
    for(i = 0; str[i]; i ++) b[2 * i + 1] = '#', b[2 * i + 2] = str[i];
    N = 2 * i + 1;
    b[0] = '$', b[N] = b[N + 1] = '#';
}
void solve()
{
    int i, j, k, id, max = 0, ans = 0;
    for(i = 1; i <= N; i ++)
    {
        p[i] = i < max ? std::min(max - i, p[2 * id - i]) : 1;    
        while(b[i + p[i]] == b[i - p[i]]) ++ p[i];
        if(i + p[i] > max) max = i + p[i], id = i;
        if(p[i] - 1 > ans)
            ans = p[i] - 1, k = (i - 1) / 2;
    }
    if(ans & 1) i = k - ans / 2, j = k + ans / 2;
    else i = k - ans / 2, j = k + ans / 2 - 1;
    for(; i <= j; i ++) printf("%c", str[i]);
    printf("\n");
}
int main()
{
    while(scanf("%s", str) == 1)
    {
        init();
        solve();    
    }
    return 0;    
}