HDU 5510 Bazinga (KMP)

题意：给定 n 个 字符串，让你找出最大的 r，使得存在一个 sl 不是sr的子串（l  < r）。

析：KMP算法，不过直接暴力就别想了，肯定TLE，所以我们考虑一下，用两个指针 l, r，如果sl 不是 sr的字串，那么们就可以更新r，继续往后，直到找到最后。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define debug puts("+++++")
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e3 + 5;
const LL mod = 2147493647;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
inline int gcd(int a, int b){  return b == 0 ? a : gcd(b, a%b); }
inline int lcm(int a, int b){  return a * b / gcd(a, b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
int f[maxn];
char s[505][maxn];

void getFail(char *P){
    int m = strlen(P);
    f[0] = f[1] = 0;
    for(int i = 1; i < m; ++i){
        int j = f[i];
        while(j && P[i] != P[j])  j = f[j];
        f[i+1] = P[i] == P[j] ? j+1 : 0;
    }
}

bool match(char *T, char *P){
    int n = strlen(T), m = strlen(P);
    getFail(P);
    int j = 0;
    for(int i = 0; i < n; ++i){
        while(j && P[j] != T[i]) j = f[j];
        if(P[j] == T[i])  ++j;
        if(j == m) return true;
    }
    return false;
}
int main(){
    int T;  cin >> T;
    for(int kase = 1; kase <= T; ++kase){
        scanf("%d", &n);
        int ans = -1;
        int l = 1, r = 2;
        for(int i = 1; i <= n; ++i)  scanf("%s", s+i);
        while(r <= n){
            while(l < r){
                if(match(s[r], s[l]))  ++l;
                else { ans = r;  break; }
            }
            ++r;
        }
        printf("Case #%d: %d\n", kase, ans);
    }
    return 0;
}