【KMP求兩子串前後最長公用子綴】HDU 2594

開始SB地暴力，就是枚舉a串，結果SB地TLE。。。

後來發現我真的很SB，完全沒有利用好kmp的優美性質，在kmp的主程式中（主程式跟預先處理程式應該分得清吧，雖然他們長得好像），每次尋找的結果就是得出b[i]對應a串中最長的前置長度j，具體是下面程式中標記部分。

#include <map>#include <set>#include <list>#include <queue>#include <deque>#include <stack>#include <string>#include <time.h>#include <cstdio>#include <math.h>#include <iomanip>#include <cstdlib>#include <limits.h>#include <string.h>#include <iostream>#include <fstream>#include <algorithm>using namespace std;#define LL long long#define MIN INT_MIN#define MAX INT_MAX#define PI acos(-1.0)#define FRE freopen("input.txt","r",stdin)#define FF freopen("output.txt","w",stdout)#define N 50005char a[N],b[N];int p[N];int n,m;void gao () {    int i,j = 0;    p[1] = 0;    for (i = 2; i <= n; i++) {        while (j > 0 && a[i] != a[j+1]) {            j = p[j];        }        if (a[i] == a[j+1]) j++;        p[i] = j;    }}void kmp () {    int i,j = 0;    for (i = 1; i <= m; i++) {        while (j > 0 && a[j+1] != b[i]) {            j = p[j];        }        if (a[j+1] == b[i]) j++;//就是這裡，得出b[i]對應a串中最長的前置長度j    }    if (!j) puts("0");    else {        for (i = 1; i <= j; i++)        printf("%c",a[i]);        printf(" %d\n",j);    }}int main () {    while (scanf("%s%s",a+1,b+1) != -1 ) {        int i,j;        n = strlen(a+1);        m = strlen(b+1);        gao();//O(n)        kmp();    }    return 0;}