标签:style blog class code java ext
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1711
题目意思:给出一条有n个数的序列a[1],a[2],......,a[n],和一条有m 个数的序列b[1],b[2],......,b[m],求出b[1],b[2],...,b[m]在序列a中完全匹配时,在序列a中的位置,如果找不到输出-1.
这几天一直在学kmp,该题算是kmp的入门题吧。有个地方要稍稍注意,代码中,主串和模式串的比较初始值为-1,-1,否则如果从0开始,会默认第一个字符是相等的!!!
1 #include <iostream> 2 #include <cstdlib> 3 #include <cstdio> 4 using namespace std; 5 6 const int N = 1e6 + 5; 7 const int M = 1e4 + 5; 8 int next[M], n, m; 9 int a[N], b[M]; 10 11 void get_next(int *next) 12 { 13 int i = 0; 14 int j = -1; 15 next[0] = -1; 16 while (i < m) 17 { 18 if (j == -1 || b[i] == b[j]) 19 { 20 i++; 21 j++; 22 next[i] = j; 23 } 24 else 25 j = next[j]; 26 } 27 } 28 29 int main() 30 { 31 int T, i, j; 32 while (scanf("%d", &T) != EOF) 33 { 34 while (T--) 35 { 36 scanf("%d%d", &n, &m); 37 for (i = 0; i < n; i++) 38 scanf("%d", &a[i]); 39 for (i = 0; i < m; i++) 40 scanf("%d", &b[i]); 41 get_next(next); 42 i = -1, j = -1; 43 int ans = -1; 44 while (i <= n && j <= m) 45 { 46 if (j == -1 || a[i] == b[j]) // 关键! 47 { 48 ++i; 49 ++j; 50 } 51 else 52 j = next[j]; 53 if (j == m) 54 { 55 ans = i - m + 1; 56 break; 57 } 58 } 59 printf("%d\n", ans); 60 } 61 } 62 return 0; 63 }
hdu 1711 Number Sequence 解题报告,布布扣,bubuko.com
标签:style blog class code java ext
原文地址:http://www.cnblogs.com/windysai/p/3717961.html