zoj 3587 Marlon's String

类型：KMP【经典】

题目：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=4629

来源： ZOJ 10th Anniversary Contest

思路：以S串为原串，T串为模式串，做两次KMP运算，分别求出在原串中以长度为i的前缀和后缀的数量

！！！如果前缀串中包含部分前缀，需要对其累加

如：S = a a a a a

              1  2 3 4 4

        T = a a a a

next[4] = 3，将前缀串长度为3的个数累加，依次计算所有含前缀串的子串

// zoj 3587 Marlon's String
// wa wa ac 30ms 2724kb
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

#define FOR(i,a,b) for(i = (a); i < (b); ++i)
#define FORE(i,a,b) for(i = (a); i <= (b); ++i)
#define FORD(i,a,b) for(i = (a); i > (b); --i)
#define FORDE(i,a,b) for(i = (a); i >= (b); --i)
#define max(a,b) ((a) > (b)) ? (a) : (b)
#define min(a,b) ((a) < (b)) ? (a) : (b)
#define CLR(a,b) memset(a,b,sizeof(a))
#define PB(x) push_back(x)

const int MAXN = 100010;

char st[MAXN], sp[MAXN];
int n, cnt, len, m;
long long p[MAXN], first[MAXN], last[MAXN];

void next() {
    int k = 0, i;
    p[1] = 0;
    FORE(i, 2, m) {
        while(k > 0 && sp[i] != sp[k + 1])
            k = p[k];
        if(sp[k + 1] == sp[i])
                k = k + 1;
        p[i] = k;
    }
}

int kmp(int sign) {
    int k = 0, i;
    FORE(i, 1, len) {
        while(k > 0 && st[i] != sp[k + 1])
            k = p[k];
        if(sp[k + 1] == st[i]) {
            (sign == 1) ? ++first[k + 1] : ++last[k + 1];
            k = k + 1;
        }
        if(k == m)
            k = p[k];
    }
    return cnt;
}

int main() {
    int i;

    scanf("%d", &n);
    while(n--) {
        scanf("%s %s", st + 1, sp + 1);
        len = strlen(st + 1);
        m = strlen(sp + 1);
        CLR(first, 0), CLR(last, 0);
        CLR(p, 0);
        next();
        kmp(1);
        FORDE(i, m, 1)
            if(p[i] != 0)
                first[p[i]] += first[i];
        char tmp;
        FORE(i, 1, len / 2)
            swap(st[i], st[len - i + 1]);
//            tmp = st[i], st[i] = st[len - i + 1], st[len - i + 1] = tmp;
        FORE(i, 1, m / 2)
            swap(sp[i], sp[m - i + 1]);
//            tmp = sp[i], sp[i] = sp[m - i + 1], sp[m - i + 1] = tmp;
        CLR(p, 0);
        next();
        kmp(0);
        FORDE(i, m, 1)
            if(p[i] != 0)
                last[p[i]] += last[i];
        long long end = 0;
        FORE(i, 1, m - 1)
                end += (long long)(first[i] * last[m - i]);
        cout<