HDU 2594 Simpsons’ Hidden Talents （kmp）

题意:

求A串的前缀和B串的后缀的lcp(最长公共前缀)

分析:

nxt数组的定义,然后把A串和B串拼起来,扫一遍nxt数组,找到第一个最大的

代码:

////  Created by TaoSama on 2015-10-30//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#include <algorithm>#include <cctype>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iomanip>#include <iostream>#include <map>#include <queue>#include <string>#include <set>#include <vector>using namespace std;#define pr(x) cout << #x << " = " << x << "  "#define prln(x) cout << #x << " = " << x << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;int n, m, nxt[N];char s[N];void getNxt() {    nxt[0] = -1;    int i = 0, j = -1;    while(i < n) {        if(j == -1 || s[i] == s[j]) nxt[++i] = ++j;        else j = nxt[j];    }}int main() {#ifdef LOCAL    freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);#endif    ios_base::sync_with_stdio(0);    while(scanf("%s", s) == 1) {        m = strlen(s);        scanf("%s", s + m);        n = strlen(s);        m = min(m, n - m);        getNxt();        int ans = 0;        for(int i = nxt[n]; i; i = nxt[i])            if(i <= m) {ans = i; break;}        if(ans) {            s[ans] = 0;            printf("%s %d\n", s, ans);        } else puts("0");    }    return 0;}