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。

View Code // HDU_3068

#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;    

}

 

View Code // URAL_1297

#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;    

}