HDU 5157 Harry and magic string (BestCoder Round #25 D) Manacher(或 Palindromic Tree) + 前缀和

题目大意：

就是现在给出一个字符串s, 长度不超过 10^5, 然后求出其中不相交的回文字串的对数

大致思路：

其实一眼看去就知道可以用Manacher处理出回文半径之后用前缀和解决 不过有想了一下Palindromic Tree的做法, 算是练习一下Palindromic Tree了

解法一：

Manacher处理出所有位置的回文半径然后计算以i位置结尾的回文串数量和以i位置开始的回文串数量, 然后其中一组乘上另外一组的前缀和加起来就是答案了

代码如下：

Result  :  Accepted     Memory  :  6016 KB     Time  :  109 ms

/*
 * Author: Gatevin
 * Created Time:  2015/3/31 10:03:33
 * File Name: Rin_Tohsaka.cpp
 */
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

#define maxn 100010

char in[maxn];
char s[maxn << 1];
int R[maxn << 1];
lint dp1[maxn], dp2[maxn], s1[maxn], s2[maxn];

void Manacher(char *s, int *R, int n)
{
    int mx = 0, p = 0;
    R[0] = 1;
    for(int i = 1; i <= n; i++)
    {
        if(mx > i) R[i] = min(R[2*p - i], mx - i);
        else R[i] = 1;
        while(s[i - R[i]] == s[i + R[i]])
            R[i]++;
        if(i + R[i] > mx)
            mx = i + R[i], p = i;
    }
    return;
}

int main()
{
    while(scanf("%s", in) != EOF)
    {
        s[0] = '@';
        int n = strlen(in);
        for(int i = 0; i < n; i++)
            s[2*i + 1] = in[i], s[2*i + 2] = '#';
        s[2*n] = '$';
        Manacher(s, R, 2*n);
        memset(dp1, 0, sizeof(dp1));
        memset(dp2, 0, sizeof(dp2));
        for(int i = 1; i < 2*n; i++)
        {
            int l = i - R[i] + 1, r = i;
            l >>= 1;
            r = (r & 1) ? r >> 1 : (r >> 1) - 1;
            if(l <= r)
                dp1[l]++, dp1[r + 1]--;
            l = i;
            r = i + R[i] - 1;
            l >>= 1;
            r = (r & 1) ? r >> 1 : (r >> 1) - 1;
            if(l <= r)
                dp2[l]++, dp2[r + 1]--;
        }
        //用s1[i]表示以第i个字符开头的回文串数量
        //用s2[i]表示以第i个字符结尾的回文串数量
        s1[0] = dp1[0], s2[0] = dp2[0];
        for(int i = 1; i < n; i++)
            s1[i] = s1[i - 1] + dp1[i], s2[i] = s2[i - 1] + dp2[i];
        //滚动数组用dp1表示s1的后缀和
        dp1[n - 1] = s1[n - 1];
        for(int i = n - 2; i >= 0; i--)
            dp1[i] = dp1[i + 1] + s1[i];
        lint ans = 0;
        for(int i = 0; i < n - 1; i++)
            ans += s2[i]*dp1[i + 1];
        printf("%I64d\n", ans);
    }
    return 0;
}

解法二：

用Palindromic Tree 来计算上面那种方法中的s1, s2 (就是把串s插入树中两次, 一次正序一次倒序就行了, 计数思想一样), 只是得到s1, s2的手段不同罢了

代码如下：

Result  :  Accepted    Memory  :  14048 KB     Time  :  109 ms

/*
 * Author: Gatevin
 * Created Time:  2015/3/31 10:34:41
 * File Name: Rin_Tohsaka.cpp
 */
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
using namespace std;
const double eps(1e-8);
typedef long long lint;

#define maxn 100010

char s[maxn];
lint dp[maxn];

struct Palindromic_Tree
{
    struct node
    {
        int next[26];
        int len;
        int sufflink;
        int cnt;
    };
    node tree[maxn];
    int L, len, suff;
    void newnode()
    {
        L++;
        for(int i = 0; i < 26; i++)
            tree[L].next[i] = -1;
        tree[L].len = tree[L].sufflink = tree[L].cnt = 0;
        return;
    }
    void init()
    {
        L = 0, suff = 2;
        newnode(), newnode();
        tree[1].len = -1; tree[1].sufflink = 1;
        tree[2].len = 0; tree[2].sufflink = 1;
        return;
    }
    bool addLetter(int pos)
    {
        int cur = suff, curlen = 0;
        int alp = s[pos] - 'a';
        while(1)
        {
            curlen = tree[cur].len;
            if(pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])
                break;
            cur = tree[cur].sufflink;
        }
        if(tree[cur].next[alp] != -1)
        {
            suff = tree[cur].next[alp];
            return false;
        }
        newnode();
        suff = L;
        tree[L].len = tree[cur].len + 2;
        tree[cur].next[alp] = L;
        if(tree[L].len == 1)
        {
            tree[L].sufflink = 2;
            tree[L].cnt = 1;
            return true;
        }
        while(1)
        {
            cur = tree[cur].sufflink;
            curlen = tree[cur].len;
            if(pos - 1 - curlen >= 0 && s[pos - 1 - curlen] == s[pos])
            {
                tree[L].sufflink = tree[cur].next[alp];
                break;
            }
        }
        tree[L].cnt = 1 + tree[tree[L].sufflink].cnt;
        return true;
    }
};

Palindromic_Tree pal;

int main()
{
    while(scanf("%s", s) != EOF)
    {
       pal.init();
       int n = strlen(s);
       for(int i = 0; i < n; i++)
       {
           pal.addLetter(i);
           dp[i] = pal.tree[pal.suff].cnt;//dp[i]表示以i位置结尾的回文串的数量
       }
       for(int i = 1; i < n; i++)
           dp[i] += dp[i - 1];//现在dp[i]表示在i位置或之前结尾的回文串数量
       reverse(s, s + n);
       pal.init();
       lint ans = 0;
       for(int i = 0; i < n; i++)
       {
           pal.addLetter(i);
           ans += pal.tree[pal.suff].cnt*dp[n - 1 - i - 1];
       }
       printf("%I64d\n", ans);
    }
    return 0;
}