hdu 5442 (ACM-ICPC2015长春网络赛F题)

题意：给出一个字符串，长度是2*10^4。将它首尾相接形成环，并在环上找一个起始点顺时针或逆时针走一圈，求字典序最大的走法，如果有多个答案则找起始点最小的，若起始点也相同则选择顺时针。

分析：后缀数组。

首先，需要补充后缀数组的基本知识的话，看这个链接。

http://www.cnblogs.com/rainydays/archive/2011/05/09/2040993.html

我利用这道题重新整理了后缀数组的模板如下：

const int MAX_LEN = 0;
//call init_RMQ(f[], n) first.
//then call query(a, b) to quest the RMQ of [a, b].
int power[30];
int st[MAX_LEN * 2][32];
int ln[MAX_LEN * 2];

//returns the index of the first minimum value in [x, y]
void init_RMQ(int f[], int n)
{
    int i, j;
    for (power[0] = 1, i = 1; i < 21; i++)
    {
        power[i] = 2 * power[i - 1];
    }
    for (i = 0; i < n; i++)
    {
        st[i][0] = i;
    }
    ln[0] = -1;
    for (int i = 1; i <= n; i++)
    {
        ln[i] = ln[i >> 1] + 1;
    }
    for (j = 1; j < ln[n]; j++)
    {
        for (i = 0; i < n; i++)
        {
            if (i + power[j - 1] - 1 >= n)
            {
                break;
            }
            //for maximum, change ">" to "<"
            //for the last, change "<" or ">" to "<=" or ">="
            if (f[st[i][j - 1]] > f[st[i + power[j - 1]][j - 1]])
            {
                st[i][j] = st[i + power[j - 1]][j - 1];
            }
            else
            {
                st[i][j] = st[i][j - 1];
            }
        }
    }
}

int query(int f[], int x, int y)
{
    if(x > y)
    {
        swap(x, y);
    }
    int k = ln[y - x + 1];
    //for maximum, change ">" to "<"
    //for the last, change "<" or ">" to "<=" or ">="
    if (f[st[x][k]] > f[st[y - power[k] + 1][k]])
        return st[y - power[k] + 1][k];
    return st[x][k];
}

//first use the constructed function
//call function lcp(l, r) to get the longest common prefix of sa[l] and sa[r]
//have access to the member of sa, myrank, height and so on
//height is the LCP of sa[i] and sa[i + 1]
class SuffixArray
{
    public:
    char* s;
    int n, sa[MAX_LEN], height[MAX_LEN], myrank[MAX_LEN];
    int tmp[MAX_LEN], top[MAX_LEN];

    SuffixArray()
    {}

    //the string and the length of the string
    SuffixArray(char* st, int len)
    {
        s = st;
        n = len + 1;
        make_sa();
        make_lcp();
    }

    void make_sa()
    {
        // O(N * log N)
        int na = (n < 256 ? 256 : n);
        memset(top, 0, na * sizeof(int));
        for (int i = 0; i < n ; i++)
            top[myrank[i] = s[i] & 0xff]++;
        for (int i = 1; i < na; i++)
            top[i] += top[i - 1];
        for (int i = 0; i < n ; i++)
            sa[--top[ myrank[i]]] = i;
        int x;
        for (int len = 1; len < n; len <<= 1)
        {
            for (int i = 0; i < n; i++)
            {
                x = sa[i] - len;
                if (x < 0)
                    x += n;
                tmp[top[myrank[x]]++] = x;
            }
            sa[tmp[top[0] = 0]] = x = 0;
            for (int i = 1; i < n; i++)
            {
                if (myrank[tmp[i]] != myrank[tmp[i-1]] ||
                        myrank[tmp[i]+len]!=myrank[tmp[i-1]+len])
                    top[++x] = i;
                sa[tmp[i]] = x;
            }
            memcpy(myrank, sa , n * sizeof(int));
            memcpy(sa , tmp, n * sizeof(int));
            if (x >= n - 1)
                break;
        }
    }

    void make_lcp()
    {
        // O(4 * N)
        int i, j, k;
        for (j = myrank[height[i = k = 0] = 0]; i < n - 1; i++, k++)
        {
            while (k >= 0 && s[i] != s[sa[j - 1] + k])
            {
                height[j - 1] = (k--);
                j = myrank[sa[j] + 1];
            }
        }
        init_RMQ(height, n - 1);
    }

    int lcp(int l, int r)
    {
        return height[query(height, l, r - 1)];
    }
};

 

先将两个原字符串拼成一个长度是2倍的字符串，便于处理越过原串末尾的情况。

运行后缀数组求顺时针解。

反转这个长字符串，再运行后缀数组，求逆时针解。

比较两个解，得出最终解。

 

下面我们来分析一下，每次运行后缀数组之后是如何求解的。

设原串长度为len。

首先从sa[len*2]开始依次向前找。即从排序最大的开始向前找。之所以排序最大的不一定是我们要的答案是因为：

1.它可能起始点在后面的附加的串上，并没有构成完整的一圈。

2.我们还要找起始点最小的答案呢。（当然，对于反转后的字符串，我们要找起始点最大的）

在第一次遇到一个起始点小于len的时候，表明这个可能就是答案。但是前面可能有起始点更小的，我们还要继续沿着sa向前找，但找的时候一定要保证字典序是最大的，所以每次向前移动一个，都要观察height(i, i+1)是否>len，如果不是，则说明字典序已经不是最大了。

特别注意，height==len的情况也是不可以的，因为两者相等很有可能表明其中一个的起始点已经等于len。当然也可以对每个都判断一下起始点位置。

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

#define d(x) 

const int MAX_N = (int)(4e4) + 100;


//call init_RMQ(f[], n) first.
//then call query(a, b) to quest the RMQ of [a, b].
int power[30];
int st[MAX_N * 2][32];
int ln[MAX_N * 2];

//returns the index of the first minimum value in [x, y]
void init_RMQ(int f[], int n)
{
    int i, j;
    for (power[0] = 1, i = 1; i < 21; i++)
    {
        power[i] = 2 * power[i - 1];
    }
    for (i = 0; i < n; i++)
    {
        st[i][0] = i;
    }
    ln[0] = -1;
    for (int i = 1; i <= n; i++)
    {
        ln[i] = ln[i >> 1] + 1;
    }
    for (j = 1; j < ln[n]; j++)
    {
        for (i = 0; i < n; i++)
        {
            if (i + power[j - 1] - 1 >= n)
            {
                break;
            }
            //for maximum, change ">" to "<"
            //for the last, change "<" or ">" to "<=" or ">="
            if (f[st[i][j - 1]] > f[st[i + power[j - 1]][j - 1]])
            {
                st[i][j] = st[i + power[j - 1]][j - 1];
            }
            else
            {
                st[i][j] = st[i][j - 1];
            }
        }
    }
}

int query(int f[], int x, int y)
{
    if(x > y)
    {
        swap(x, y);
    }
    int k = ln[y - x + 1];
    //for maximum, change ">" to "<"
    //for the last, change "<" or ">" to "<=" or ">="
    if (f[st[x][k]] > f[st[y - power[k] + 1][k]])
        return st[y - power[k] + 1][k];
    return st[x][k];
}

//first use the constructed function
//call function lcp(l, r) to get the longest common prefix of sa[l] and sa[r]
//have access to the member of sa, myrank, height and so on
//height is the LCP of sa[i] and sa[i + 1]
class SuffixArray
{
    public:
    char* s;
    int n, sa[MAX_N], height[MAX_N], myrank[MAX_N];
    int tmp[MAX_N], top[MAX_N];

    SuffixArray()
    {}

    //the string and the length of the string
    SuffixArray(char* st, int len)
    {
        s = st;
        n = len + 1;
        make_sa();
        make_lcp();
    }

    void make_sa()
    {
        // O(N * log N)
        int na = (n < 256 ? 256 : n);
        memset(top, 0, na * sizeof(int));
        for (int i = 0; i < n ; i++)
            top[myrank[i] = s[i] & 0xff]++;
        for (int i = 1; i < na; i++)
            top[i] += top[i - 1];
        for (int i = 0; i < n ; i++)
            sa[--top[ myrank[i]]] = i;
        int x;
        for (int len = 1; len < n; len <<= 1)
        {
            for (int i = 0; i < n; i++)
            {
                x = sa[i] - len;
                if (x < 0)
                    x += n;
                tmp[top[myrank[x]]++] = x;
            }
            sa[tmp[top[0] = 0]] = x = 0;
            for (int i = 1; i < n; i++)
            {
                if (myrank[tmp[i]] != myrank[tmp[i-1]] ||
                        myrank[tmp[i]+len]!=myrank[tmp[i-1]+len])
                    top[++x] = i;
                sa[tmp[i]] = x;
            }
            memcpy(myrank, sa , n * sizeof(int));
            memcpy(sa , tmp, n * sizeof(int));
            if (x >= n - 1)
                break;
        }
    }

    void make_lcp()
    {
        // O(4 * N)
        int i, j, k;
        for (j = myrank[height[i = k = 0] = 0]; i < n - 1; i++, k++)
        {
            while (k >= 0 && s[i] != s[sa[j - 1] + k])
            {
                height[j - 1] = (k--);
                j = myrank[sa[j] + 1];
            }
        }
        init_RMQ(height, n - 1);
    }

    int lcp(int l, int r)
    {
        return height[query(height, l, r - 1)];
    }
};

SuffixArray suffix;
int len;
char ans1[MAX_N], ans2[MAX_N];
int pos1, pos2;
char s[MAX_N];

int work1(char* ans)
{
    int p = len * 2;
    while (suffix.sa[p] >= len)
        p--;
    for (int i = p - 1; i >= 0; i--)
    {
        if (suffix.lcp(i, i + 1) <= len)
            break;
        if (suffix.sa[i] < suffix.sa[p])
        {
            p = i;
        }
    }
    memcpy(ans, s + suffix.sa[p], len);
    ans[len] = 0;
    return suffix.sa[p];
}

int work2(char* ans)
{
    int p = len * 2;
    while (suffix.sa[p] >= len)
        p--;
    for (int i = p - 1; i >= 0; i--)
    {
        if (suffix.lcp(i, i + 1) <= len)
            break;
        if (suffix.sa[i] > suffix.sa[p])
        {
            p = i;
        }
    }
    memcpy(ans, s + suffix.sa[p], len);
    ans[len] = 0;
    return suffix.sa[p];
}

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%d", &len);
        scanf("%s", s);
        for (int i = 0; i < len; i++)
        {
            s[len + i] = s[i];
        }
        suffix = SuffixArray(s, len * 2);
        pos1 = work1(ans1);
        reverse(s, s + len * 2);
        suffix = SuffixArray(s, len * 2);
        pos2 = work2(ans2);
        pos2 = len - 1 - pos2;

        if (strcmp(ans1, ans2) > 0)
        {
            printf("%d 0\n", pos1 + 1);
            continue;
        }else if (strcmp(ans1, ans2) < 0)
        {
            printf("%d 1\n", pos2 + 1);
            continue;
        }else if (pos1 <= pos2)
        {
            printf("%d 0\n", pos1 + 1);
            continue;
        }else
        {
            printf("%d 1\n", pos2 + 1);
        }
    }
    return 0;
}