(KMP 1.1)hdu 1711 Number Sequence(KMPの簡単な応用--patternがtextの中で初めて現れる位置を求める)
3127 ワード
タイトル:
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
Sample Output
Source
HDU 2007-Spring Programming Contest
Recommend
lcy | We have carefully selected several similar problems for you: 1358 3336 1686 3746 2222
テーマ分析:
KMP.簡単な問題.
コードは次のとおりです.
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
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
Source
HDU 2007-Spring Programming Contest
Recommend
lcy | We have carefully selected several similar problems for you: 1358 3336 1686 3746 2222
テーマ分析:
KMP.簡単な問題.
コードは次のとおりです.
#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 1000001;
int n;//
int m;//
int text[maxn];//
int pattern[maxn];//
int nnext[maxn];//next . next
/*O(m) next */
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)
*
*
*/
int kmp() {
int j = 0;/* */
for (int i = 0; i < n; i++) {/* */
while (j && pattern[j] != text[i])/* , , j = 0*/
j = nnext[j];
if (pattern[j] == text[i])/* */
j++;
if (j == m) {/* */
// printf("%d
" , i-m+2);/* , */
return i - m + 2;// 1
}
}
// printf("-1
");
return -1; //
}
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
", kmp());
}
return 0;
}