CodeForces 113B Petr#

题目链接：http://codeforces.com/problemset/problem/113/B

题目大意：

多组数据
每组给定3个字符串T，Sbeg，Sed，求字符串T中有多少子串是以Sbeg开头，Sed结尾的

分析：

　　难点在哈希函数的编写，如果直接存string会爆内存，不能用STL自带哈希函数，用了会Wa。

代码如下：

#include <bits/stdc++.h>
using namespace std;
 
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
 
#define pii pair<int,int> 
#define piii pair<pair<int,int>,int> 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
 
inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
     
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 
 
inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
const LL mod = 1e9 + 7;
const int maxN = 2000 + 7;
 
string T, Sbeg, Sed;
unordered_set< LL > sll;
int beg[maxN], begLen; // 记录 Sbeg出现的位置 
int ed[maxN], edLen; // 记录 Sed出现的位置 
 
// h为T的后缀哈希数组 
// h[i]表示T从i位置开始的后缀的哈希值 
// h[i] = T[i] + T[i+1]*key + T[i+2]*key^2 + ……+ T[i+len-1-i]*key^len-1-i
// xp为基数数组
// xp[i] = key^i
LL xp[maxN], h[maxN];
const LL key = 1e9 + 7;
 
// 求起点为s，长为len的子串的哈希值 
// Hash(i, len) = T[i] + T[i+1]*key + T[i+2]*key^2 + ……+ T[i+len-1]*key^len-1
LL Hash(int s, int len) {
    return h[s] - h[s + len] * xp[len];
}
  
void HashInit(const char* s, LL* h, int len) {
    xp[0] = 1;
    For(i, 1, maxN - 1) xp[i] = xp[i - 1] * key;
         
    h[len] = 0;
    rFor(i, len - 1, 0) h[i] = h[i + 1] * key + s[i];        
}
 
int main(){
    while(cin >> T >> Sbeg >> Sed) {
        HashInit(T.c_str(), h, (int)T.size());
         
        int ans = 0;
        sll.clear();
        begLen = edLen = 0;
         
        int p = 0;
        while(p < T.size()) {
            int t = T.find(Sbeg.c_str(), p);
            if(t == string::npos) break;
            beg[begLen++] = t;
            p = t + 1;
        }
         
        p = 0;
        while(p < T.size()) {
            int t = T.find(Sed.c_str(), p);
            if(t == string::npos) break;
            ed[edLen++] = t;
            p = t + 1;
        }
         
        int i = 0, j = 0;
        while(i < begLen && j < edLen) {
            if(beg[i] <= ed[j]) {
                For(k, j, edLen - 1) {
                    if(beg[i] + Sbeg.size() > ed[k] + Sed.size()) continue;
                    sll.insert(Hash(beg[i], ed[k] + Sed.size() - beg[i]));
                }
                ++i;
            }
            else ++j;
        }
         
        cout << sll.size() << endl;
    }
    return 0;
}