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`
题目分析：
字符串匹配的问题，kmp算法的应用。定义主串s和字串t，定位字串t就是要在主串s中找到一个与字串t完全相等的子串，返回第一次找到的位置的下标。
KMP算法
#includeconst int maxn = 10000 + 10, maxm = 1000000 + 10;int n, m;int p[maxn], t[maxm], next[maxn]; //定义模式串，文本串，next数组void getNext() //O(m)复杂度求Next数组{ int i = 0, j = 0, k = -1; next[0] = -1; while(j < m) { if(k == -1 || p[k] == p[j]) next[++j] = ++k; else k = next[k]; } }int kmp(){ int j = 0; //初始化模式串的前一个位置 getNext(); //生成next数组 for(int i = 0; i < n; i++) //遍历文本串 { while(j && p[j] != t[i]) j = next[j];//持续走直到可以匹配 if(p[j] == t[i]) j++; //匹配成功继续下一个位置（j先加1 此时j的位置所对应的值在i的位置所对应的后面） //if(j==m)说明子串已经匹配完了（子串的长度是m，m-1是子串的最后一个字符） if(j == m) return i-m+2; //找到后返回第一个匹配的位置,因为返回的是逻辑下标（下标从1开始所以是i-m+2，不信自己在纸上画画 （看下图 ）） } return -1; //找不到返回-1}int main(){ int T; scanf("%d", &T); while(T--) { scanf("%d%d", &n, &m); for(int i = 0; i < n; i++) scanf("%d", t+i); for(int i = 0; i < m; i++) scanf("%d", p+i); printf("%d\n", kmp()); } return 0;}
解释为什么返回 i-m+2