[HDOJ1711]Number Sequence

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題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=1711

 

 

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15683    Accepted Submission(s): 6898

Problem DescriptionGiven 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. 

 

InputThe 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]. 

 

OutputFor each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead. 

 

Sample Input213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1 

 

Sample Output6-1 

 

初學KMP，水一發模版題

 

 

 1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <iostream> 6 #include <cmath> 7 #include <queue> 8 #include <map> 9 #include <stack>10 #include <list>11 #include <vector>12 13 using namespace std;14 const int maxn = 1000010;15 int na, nb;16 int a[maxn];17 int b[maxn];18 int pre[maxn];19 20 void getpre(int *b, int *pre) {21     int j, k;22     pre[0] = -1;23     j = 0;24     k = -1;25     while(j < nb - 1) {26         if(k == -1 || b[j] == b[k]) {//匹配27             j++;28             k++;29             pre[j] = k;30         }31         else {  //b[j] != b[k]32             k = pre[k];33         }34     }35 }36 37 int kmp() {38     int i = 0;39     int j = 0;40     getpre(b, pre);41     while(i < na) {42         if(j == -1 || a[i] == b[j]) {43             i++;44             j++;45         }46         else {47             j = pre[j];48         }49         if(j == nb) {50             return i - nb + 1;51         }52     }53     return -1;54 }55 56 int main() {57     int T;58     scanf("%d", &T);59     while(T--) {60         scanf("%d %d", &na, &nb);61         for(int i = 0; i < na; i++) {62             scanf("%d", &a[i]);63         }64         for(int i = 0; i < nb; i++) {65             scanf("%d", &b[i]);66         }67         printf("%d\n", kmp());68     }69     return 0;70 }

 

本題還可以用hash做，效率與kmp差距不大。

 

 

 1 #include <cstdio> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <iostream> 6 #include <cmath> 7 #include <queue> 8 #include <map> 9 #include <set>10 #include <stack>11 #include <list>12 #include <vector>13 14 using namespace std;15 16 typedef unsigned long long ull;17 const int B = 100007;18 const int maxn = 1000010;19 int a[maxn], b[maxn];20 int na, nb;21 22 ull quickmul(int x, int n) {23     ull ans = 1;24     ull t = x;25     while(n) {26         if(n & 1) {27             ans = (ans * t);28         }29         t = t * t;30         n >>= 1;31     }32     return ans;33 }34 35 int contain() {36     if(na > nb) {37         return false;38     }39     ull t = quickmul(B, na);40     ull ah = 0, bh = 0;41     for(int i = 0; i < na; i++) {42         ah = ah * B + a[i];43     }44     for(int i = 0; i < na; i++) {45         bh = bh * B + b[i];46     }47     for(int i = 0; i + na <= nb; i++) {48         if(ah == bh) {49             return i + 1;50         }51         if(i + na < nb) {52             bh = bh * B + b[i+na] - b[i] * t;53         }54     }55     return -1;56 }57 int main() {58     // freopen("in", "r", stdin);59     int T;60     scanf("%d", &T);61     while(T--) {62         scanf("%d %d", &nb, &na);63         for(int i = 0; i < nb; i++) {64             scanf("%d", &b[i]);65         }66         for(int i = 0; i < na; i++) {67             scanf("%d", &a[i]);68         }69         printf("%d\n", contain());70     }71     return 0;72 }

 

[HDOJ1711]Number Sequence