Xi, a developmental biologist is working on developmental distances of chromosomes. A chromosome, in the Xi‘s simplistic view, is a permutation from n genes numbered 1 to n. Xi is working on an evolutionary distance metric between two chromosomes. In Xi‘s theory of evolution any subset of genes lying together in both chromosomes is a positive witness for chromosomes to be similar.
A positive witness is a pair of sequence of the same length A and A‘, where A is a consecutive subsequence of the first chromosome, A‘ is a consecutive subsequence of the second chromosome, and A is a permutation ofA‘. The goal is to count the number of positive witnesses of two given chromosomes that have a length greater than one.
Input
There are several test case in the input. Each test case starts with a line containing the number of genes(2n
3000).
The next two lines contain the two chromosomes, each as a list of positive integers. The input terminates with a line containing ``0‘‘ which should not be processed as a test case.
Output
For each test case, output a single line containing the number of positive witness for two chromosomes to be similar.
Sample Input
4 3 2 1 4 1 2 4 3 5 3 2 1 5 4 3 2 1 5 4 0
Sample Output
3 10
题意：说白了就是求两个集合有几个相同的连续子集。
思路：记录下第二个集合每个数的位置。然后去枚举第一个集合。把子集内的数字位置的Max和Min记录下来，然后如果Max - Min + 1 == len说明集合长度相同。在记录过程中，利用了类似前缀和的思想，复杂度为O(n^2)。
代码：
#include <stdio.h>
#include <string.h>
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
const int N = 3005;
int n, a[N], vis[N];
void init() {
int num;
for (int i = 0; i < n; i++)
scanf("%d", &a[i]);
for (int j = 0; j < n; j++) {
scanf("%d", &num);
vis[num] = j;
}
}
int solve() {
int ans = 0;
for (int i = 0; i < n; i++) {
int Max = vis[a[i]], Min = vis[a[i]], len = 1;
for (int j = i + 1; j < n; j++) {
Max = max(Max, vis[a[j]]);
Min = min(Min, vis[a[j]]);
len++;
if (Max - Min + 1 == len) {
ans++;
}
}
}
return ans;
}
int main() {
while (~scanf("%d", &n) && n) {
init();
printf("%d\n", solve());
}
return 0;
}