[The Problem to Slow Down You] 后缀自动机 + 马拉车做法

链接的G题:   http://codeforces.com/gym/100548                                 

1. 由于不会回文树， 看到题目感觉很SAM

2. 仔细一想可以发现 SAM 中一个节点所代表的字符串最多只有一个是回文串

以同一个字母结尾的不同回文串 出现的位置不可能完全相同

说明一个长为n的串种出现的回文子串种类不超过n种

3.在新建一个结点的时候如何判断——该节点所存储的字符串中是否有回文串呢?

如果有的话，必然是以当前字符结尾的最长的那个回文串

Manacher处理一下 最长的有多长。 然后比较一下是否大于等于minlen[cur] 就行了

4.在匹配到SAM上一个节点时具体要加多少呢?

假设当前匹配到的字符串长度为len

那要加的就是比len短的回文串在模式串中出现的次数

具体就是suffix_link上去找, 记得乘上其出现的次数

5. 顺便回头看了一下， 好像回文树能做的， SAM + manacher 都能做?

#include <iostream>#include <cstring>#include <cstdio>#include <vector>#include <set>#include <map>#include <queue>#include <algorithm>#include <stack>#include <cctype>#include <cmath>#include <vector>#include <sstream>#include <bitset>#include <deque>#include <iomanip>using namespace std;#define pr(x) cout << #x << " = " << x << endl;#define bug cout << "bugbug" << endl;#define ppr(x, y) printf("(%d, %d)\n", x, y);#define MST(a,b) memset(a,b,sizeof(a))#define CLR(a) MST(a,0)#define SQR(a) ((a)*(a))#define PCUT puts("\n---------------")typedef long long ll;typedef double DBL;typedef pair<int, int> P;typedef unsigned int uint;const int MOD = 1e9 + 7;const int inf = 0x3f3f3f3f;const ll INF = 0x3f3f3f3f3f3f3f3f;const int maxn = 4e5 + 4;const int maxm = 1e3 + 4;const double pi = acos(-1.0);int m[maxn], M[maxn], Temp[maxn];char buf[maxn];void manacher(char* s){int j = 1;buf[0] = '#';for (int i = 0; s[i]; ++i){buf[j++] = '$';buf[j++] = s[i];}buf[j++] = '$';buf[j++] = 0;int center = -1, maxv = -1;for (int i = 1; buf[i]; ++i){m[i] = i <= maxv ? min(m[2*center-i], maxv - i + 1) : 1;while(buf[i-m[i]] == buf[i+m[i]]) m[i]++;if (i + m[i] - 1 > maxv){maxv = i + m[i] - 1;center = i;}}int lst = 0;for (int i = 1; buf[i]; ++i)if (i + m[i] - 1 > lst) while (i + m[i] - 1 > lst){lst++;Temp[lst] = (lst - i) * 2 + 1;}for (int i = 2; buf[i]; i += 2){M[i/2-1] = (Temp[i] + 1) / 2;}return;} stack<int> sta;struct SuffixAutoMachine{int trans[maxn][26], maxlen[maxn], link[maxn], cnt, lst;//写成了我手上AC自动机模板那个风格 int times[maxn], in[maxn];ll plus[maxn], len_p[maxn];//当前节点的回文串长度 inline int id(char x){return x - 'a';}void newNode(){cnt++;memset(trans[cnt], -1, sizeof trans[cnt]);link[cnt] = len_p[cnt] = -1;times[cnt] = plus[cnt] = 0;return; }void insert(int id){newNode();maxlen[cnt] = maxlen[lst] + 1;times[cnt] = 1; int u = lst;while(u != -1 && trans[u][id] == -1){trans[u][id] = cnt;u = link[u];} if (u == -1){link[cnt] = 0;lst = cnt;return;}int q = trans[u][id];if (maxlen[q] == maxlen[u] + 1){link[cnt] = q;lst = cnt;return;}int cur = cnt, sq = cnt+1;newNode();maxlen[sq] = maxlen[u] + 1;memcpy(trans[sq], trans[q], sizeof trans[q]);link[sq] = link[q];link[q] = sq;link[cur] = sq;lst = cur;if (len_p[q] != -1 && len_p[q] <= maxlen[sq]){len_p[sq] = len_p[q];len_p[q] = -1;}while(u != -1 && trans[u][id] == q){trans[u][id] = sq;u = link[u];}return;}void init(){cnt = -1;newNode();maxlen[0] = 0;lst = 0;}void construct(char* s){init();for (int i = 0; s[i]; ++i){insert(id(s[i]));int bef = link[lst];//cout << s[i] << ' ' << M[i] << ' ' << maxlen[bef] << endl;if (M[i] > maxlen[bef]) len_p[lst] = M[i];}while(sta.size()) sta.pop();memset(in, 0, sizeof in);for (int i = 1; i <= cnt; ++i) in[link[i]]++;queue<int> q;for (int i = 1; i <= cnt; ++i) if (!in[i]) q.push(i);while(q.size()){int top = q.front(); q.pop(); sta.push(top);if (top == 0) continue;times[link[top]] += times[top];if (--in[link[top]] == 0) q.push(link[top]);} while(sta.size()){int top = sta.top(); sta.pop();if (top == 0) continue;plus[top] = plus[link[top]];if (len_p[top] != -1) plus[top] += times[top];}}ll match(char *s){ll sum = 0;int cur = 0, len = 0;for (int i = 0; s[i]; ++i){int j = id(s[i]);while(cur && trans[cur][j] == -1) cur = link[cur], len = maxlen[cur];if (trans[cur][j] != -1){cur = trans[cur][j];//cout << i << ' ' << maxlen[cur] << ' ' << plus[cur] << endl;len++;//cout << len_p[cur] << endl;if (len_p[cur] != -1 && len_p[cur] <= len) sum += plus[cur];else sum += plus[link[cur]];}}return sum;}}SAM;char a[maxn], b[maxn];int main(){//必须编译过才能交int ik, i, j, k, kase;//SAM要求有两倍字符串长度的节点。 //匹配时 --i 和 cur == 0两个需要注意的地方 scanf("%d", &kase);for (ik = 1; ik <= kase; ++ik){scanf("%s%s", a, b);manacher(a);SAM.construct(a);printf("Case #%d: %I64d\n", ik, SAM.match(b));}return 0;}/*10ewewewweeeewwwwqewewewweeee*/