标签:1.0 lcm tle case ons names inf bug strlen
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; }
标签:1.0 lcm tle case ons names inf bug strlen
原文地址:http://www.cnblogs.com/dwtfukgv/p/6024811.html