【CF】86 B. Petr#

误以为是求满足条件的substring总数（解法是KMP分别以Sbeg和Send作为模式串求解满足条件的position,然后O(n^2)或者O(nlgn)求解）。
后来发现是求set(all valid substring), O(n^2)遍历起始和终止位置并利用set肯定超时。因此，利用字符串hash求解。

 /* 113B */
 #include <iostream>
 #include <string>
 #include <map>
 #include <queue>
 #include <set>
 #include <stack>
 #include <vector>
 #include <deque>
 #include <algorithm>
 #include <cstdio>
 #include <cmath>
 #include <ctime>
 #include <cstring>
 #include <climits>
 #include <cctype>
 #include <cassert>
 #include <functional>
 #include <iterator>
 #include <iomanip>
 using namespace std;
 //#pragma comment(linker,"/STACK:102400000,1024000")

 #define sti                set<int>
 #define stpii            set<pair<int, int> >
 #define mpii            map<int,int>
 #define vi                vector<int>
 #define pii                pair<int,int>
 #define vpii            vector<pair<int,int> >
 #define rep(i, a, n)     for (int i=a;i<n;++i)
 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 #define clr                clear
 #define pb                 push_back
 #define mp                 make_pair
 #define fir                first
 #define sec                second
 #define all(x)             (x).begin(),(x).end()
 #define SZ(x)             ((int)(x).size())
 #define lson            l, mid, rt<<1
 #define rson            mid+1, r, rt<<1|1

 const int prime = ;
 const int maxn = ;
 char s[maxn];
 char s1[maxn];
 char s2[maxn];
 int nxt1[maxn];
 int nxt2[maxn];
 bool valid1[maxn];
 bool valid2[maxn];
 __int64 Pow[maxn];
 __int64 hash[maxn];

 void getNext(char *s, int nxt[]) {
     int slen = strlen(s);
     int i = , j = -;

     nxt[] = -;
     while (i < slen) {
         if (j==- || s[i]==s[j]) {
             ++i;
             ++j;
             nxt[i] = j;
         } else {
             j = nxt[j];
         }
     }
 }

 void kmp(char *s, char *d, int nxt[], bool valid[]) {
     int slen = strlen(s);
     int dlen = strlen(d);
     int i = , j = ;

     while (i < dlen) {
         if (d[i] == s[j]) {
             ++i;
             ++j;
         } else {
             j = nxt[j];
             if (j == -) {
                 j = ;
                 ++i;
             }
         }
         if (j == slen) {
             valid[i-slen] = true;
             j = nxt[j];
         }
     }
 }

 void init() {
     Pow[] = ;
     rep(i, , maxn)
         Pow[i] = Pow[i-] * prime;

     int slen = strlen(s);
     hash[] = s[] - 'a' + ;
     rep(i, , slen)
         hash[i] = hash[i-] * prime + s[i]-'a'+;
 }

 __int64 getHash(int b, int e) {
     return hash[e] - (b== ? 0LL : hash[b-]) * Pow[e-b+];
 }

 int main() {
     ios::sync_with_stdio(false);
     #ifndef ONLINE_JUDGE
         freopen("data.in", "r", stdin);
         freopen("data.out", "w", stdout);
     #endif

     scanf("%s %s %s", s, s1, s2);

     init();
     getNext(s1, nxt1);
     getNext(s2, nxt2);

     int i, j, k;
     int slen = strlen(s);
     int slen1 = strlen(s1);
     int slen2 = strlen(s2);

     kmp(s1, s, nxt1, valid1);
     kmp(s2, s, nxt2, valid2);

     vector<__int64> vc;

     for (i=; i<slen; ++i) {
         if (!valid1[i])
             continue;
         for (j=i; j<slen; ++j) {
             if (valid2[j] && i+slen1<=j+slen2) {
                 vc.pb(getHash(i, j+slen2-));
             }
         }
     }

     sort(all(vc));
     int sz = SZ(vc);
     int ans = ;

     for (i=sz-; i>=; --i) {
         if (i== || vc[i]!=vc[i-])
             ++ans;
     }
     printf("%d\n", ans);

     #ifndef ONLINE_JUDGE
         printf("time = %d.\n", (int)clock());
     #endif

     return ;
 }