Simpsons’ Hidden Talents（字符串-KMP算法）

Simpsons’ Hidden Talents（字符串-KMP算法）

ｊｕｄｇｅ：ＨＤＵＯＪ　２５９４
Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 32768/32768 K (Java/Others)
ｓｏｕｒｃｅ：HDU 2010-05 Programming Contest

Problem Description

Homer: Marge, I just figured out a way to discover some of the talents we weren’t aware we had.
Marge: Yeah, what is it?
ＨＤＵＯＪ　２５９４

Input

ＨＤＵＯＪ　２５９４

Output

ＨＤＵＯＪ　２５９４

Sample Input

clinton
homer
riemann
marjorie

Sample Output

0
rie 3

题意

给你两个字符串a和b，其中a的前缀可能和b的后缀有重合部分，让你求出重合部分并输出。

先在a的末尾加一个原来没有的字符（如’ ‘或者’#’，加完不要忘记长度也要更新），然后把b复制到a的末尾，跑一波KMP求next数组，最后看Next[alen + blen]就完事了

代码

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#define _for(i, a) for(int i = 0; i < (a); i++)
#define _rep(i, a, b) for(int i = (a); i <= (b); i++)
typedef long long ll;
const int maxn = 100005;
const int inf = 0x3f3f3f3f;
using namespace std;

int Next[maxn];
char a[maxn], b[maxn];

void GetNext(char p[], int len) {
	int plen = strlen(p);
	Next[0] = -1;
	int j = 0, k = -1;
	while (j < plen) {
		if (k == -1 || p[j] == p[k]) {
			k++;
			j++;
			Next[j] = k;
		}
		else {
			k = Next[k];
		}
	}
}

int main() {
	//freopen("in.txt", "r", stdin);
	ios::sync_with_stdio(false);
	while (cin >> a >> b) {
		int alen = strlen(a),
			blen = strlen(b);
		a[alen++] = ' ';
		_for(i, blen) {
			a[alen + i] = b[i];
		}
		GetNext(a, alen + blen);
		int ans = Next[alen + blen];
		_for(i, ans) printf("%c", a[i]);
		if (ans) printf(" ");
		printf("%d\n", ans);
	}
	return 0;
}