(KMP 1.1)hdu 1711 Number Sequence(KMP的簡單應用——求pattern在text中第一次出現的位置)


題目:

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 12902    Accepted Submission(s): 5845


Problem 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
213 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
 
Source HDU 2007-Spring Programming Contest  
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題目分析:

               KMP。簡單題。



代碼如下:


#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>using namespace std;const int maxn = 1000001;int n;//文本串的長度int m;//目標串的長度int text[maxn];//文本串int pattern[maxn];//模式串int nnext[maxn];//next數組.直接起next可能會跟系統中預定的重名/*O(m)的時間求next數組*/void get_next() {nnext[0] = nnext[1] = 0;for (int i = 1; i < m; i++) {int j = nnext[i];while (j && pattern[i] != pattern[j])j = nnext[j];nnext[i + 1] = pattern[i] == pattern[j] ? j + 1 : 0;}}/*o(n)的時間進行匹配 * * 返回第一次匹配的位置 */int kmp() {int j = 0;/*初始化在模式串的第一個位置*/for (int i = 0; i < n; i++) {/*遍歷整個文本串*/while (j && pattern[j] != text[i])/*順着失配邊走,直到可以匹配,最壞得到情況是j = 0*/j = nnext[j];if (pattern[j] == text[i])/*如果匹配成功繼續下一個位置*/j++;if (j == m) {/*如果找到了直接輸出*///         printf("%d\n" , i-m+2);/*輸出在文本串中第一個匹配的位置,不是下標*/return i - m + 2;//返回的位置從1開始算}}//    printf("-1\n");return -1; //表示沒有找到匹配的位置}int main() {int t;scanf("%d", &t);while (t--) {scanf("%d%d", &n, &m);int i;for (i = 0; i < n; ++i) {scanf("%d", &text[i]);}for (i = 0; i < m; ++i) {scanf("%d", &pattern[i]);}get_next();printf("%d\n", kmp());}return 0;}






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