hdu 2594 Simpsons’ Hidden Talents

hdu 2594 Simpsons’ Hidden Talents

求第一个字符串StrA的前缀和第二个字符串strB的后缀的最大长度.

只要把strA当做模式串去匹配strB就可以啦， 不过还要注意被匹配串strB的长度哦。

例如 strA = 'abcabcabcabc', strB = 'abcabcabcabcabc' 答案却是 abc 3...

#include <stdio.h>
#include <string.h>

#define MAX 50005

char name[MAX], talentName[MAX];
int next[MAX];

void getNext(char* str, int* next) {
	int len;
	int j, k;
	
	j = 0;
	k = -1;
	next[0] = k;
	len = strlen(str) - 1;

	while (j < len) {
		if (k == -1 || str[k] == str[j]) {
			++k, ++j;

			if (str[k] == str[j]) {
				next[j] = next[k];
			} else {
				next[j] = k;
			}
		} else {
			k = next[k];
		}
	}
}

int kmp(char* strA, char* strB) {
	int lenA, lenB;
	int i, j;

	i = j = 0;
	lenA = strlen(strA);
	lenB = strlen(strB);
	getNext(strB, next);

	if (lenA > lenB) {
		i = lenA - lenB;
	}

	while (i < lenA) {
		if (j == -1 || strA[i] == strB[j]) {
			++j, ++i;
		} else {
			j = next[j];
		}
	}

	return j;
}

int main() {
	int matchNum;
	int i;

	while (scanf("%s%s", name, talentName) != EOF) {
		matchNum = kmp(talentName, name);
		if (matchNum == 0) {
			printf("0\n");
		} else {
			for (i = 0; i < matchNum; i++) {
				printf("%c", name[i]);
			}
			printf(" %d\n", matchNum);
		}
	}

	return 0;
}