HDU - 1711 Number Sequence KMP字符串匹配

HDU - 1711 Number Sequence Time Limit: 5000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u Submit Status Description Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.    Input The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].    Output For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.    Sample Input 2 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 1 3 13 5 1 2 1 2 3 1 2 3 1 3 2 1 2 1 2 3 2 1   Sample Output 6 -1 建议使用第一类型代码，因为第二类型代码存在错误，面对一些特殊情况，他的结果是不正确的  第一类型代码： /* Problem : 1711 ( Number Sequence ) Judge Status : Accepted RunId : 14382956 Language : G++ Author : 24862486 Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta */ #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; const int MAXN = 10000 + 5; const int MAXM = 1000000 + 5; int T, N, M, a[MAXM], b[MAXN], nexts[MAXN]; void Get_Next() { int i = 0, j = -1; nexts[0] = -1; while(i < M - 1) { if(j == -1 || b[j] == b[i]) { j ++; i ++; if(b[j] == b[i]) { nexts[i] = nexts[j]; } else { nexts[i] = j; } } else { j = nexts[j]; } } } int Get_KMP() { int i = 0, j = 0; while(i < N && j < M) { if(j == -1 || a[i] == b[j]) { i ++; j ++; } else { j = nexts[j];//失配匹配 } } if(j == M) { return i - j + 1; } return -1; } int main() { scanf("%d", &T); while(T --) { scanf("%d %d", &N, &M); for(int i = 0; i < N; i ++) { scanf("%d", &a[i]); } for(int i = 0; i < M; i ++) { scanf("%d", &b[i]); } Get_Next(); int cnt = Get_KMP(); printf("%d\n", cnt); } return 0; } 第二中类型：效率较低，但是代码简单，便于记忆以及理解 /* Problem : 1711 ( Number Sequence ) Judge Status : Accepted RunId : 14382956 Language : G++ Author : 24862486 Code Render Status : Rendered By HDOJ G++ Code Render Version 0.01 Beta */ #include <cstdio> #include <cstring> #include <algorithm> #include <vector> #include <queue> using namespace std; const int MAXN = 10000 + 5; const int MAXM = 1000000 + 5; int T, N, M, a[MAXM], b[MAXN], nexts[MAXN]; void Get_Next() { int i = 0, j = -1; nexts[0] = -1; while(i < M) { if(j == -1 || b[j] == b[i]) { nexts[++ i] = ++ j; } else { j = nexts[j]; } } } int Get_KMP() { int i = 0, j = 0; while(i < N && j < M) { if(j == -1 || a[i] == b[j]) { i ++; j ++; } else { j = nexts[j];//失配匹配 } } if(j == M) { return i - j + 1; } return -1; } int main() { scanf("%d", &T); while(T --) { scanf("%d %d", &N, &M); for(int i = 0; i < N; i ++) { scanf("%d", &a[i]); } for(int i = 0; i < M; i ++) { scanf("%d", &b[i]); } Get_Next(); int cnt = Get_KMP(); printf("%d\n", cnt); } return 0; }