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 100010char 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 100010char 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;}