UVaLive LA 5052 UVa 1481 - Genome Evolution （很巧妙的思维）

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Xi, a developmental biologist is working on developmental distances of chromosomes. A chromosome, in the Xi's simplistic view, is a permutation fromn 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 of A'. 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(2 $\le$ n $\le$ 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 43 2 1 5 40

Sample Output

3 10

题意：

给出1-n的两个排列A和B，统计有多少二元组(A',B')满足：A' B' 分别是A B的子序列，且A' B'包含的整数集完全相同 A' B'均应至少包含两个元素

思路：

我没有想出来，看了网上的题解，发现智商是硬伤啊

补充一下大白上面题意 A B序列均没有重复元素所以才可以这有做

直接枚举A的所有子序列O（n^2）同时A的子序列subA的元素在B中出现的位置的最大值和最小值

如果位置的最大值和最小值之差+1==len(subA) 那么这个就是一个解因为没有重复元素！

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <cstring>#include <string>#include <map>#include <cmath>#include <queue>#include <set>using namespace std;//#define WIN#ifdef WINtypedef __int64 LL;#define iform "%I64d"#define oform "%I64d\n"#define oform1 "%I64d"#elsetypedef long long LL;#define iform "%lld"#define oform "%lld\n"#define oform1 "%lld"#endif#define S64I(a) scanf(iform, &(a))#define P64I(a) printf(oform, (a))#define P64I1(a) printf(oform1, (a))#define REP(i, n) for(int (i)=0; (i)<n; (i)++)#define REP1(i, n) for(int (i)=1; (i)<=(n); (i)++)#define FOR(i, s, t) for(int (i)=(s); (i)<=(t); (i)++)const int INF = 0x3f3f3f3f;const double eps = 10e-9;const double PI = (4.0*atan(1.0));const int maxn = 3000 + 20;int A[maxn];int B[maxn];int main() {    int n;    while(scanf("%d", &n) != EOF && n) {        for(int i=1; i<=n; i++) scanf("%d", &A[i]);        for(int i=1; i<=n; i++) {            int t;            scanf("%d", &t);            B[t] = i;        }        int ans = 0;        for(int i=1; i<n; i++) {            int minp = B[A[i]];            int maxp = B[A[i]];            for(int j=i+1; j<=n; j++) {                int p = B[A[j]];                minp = min(minp, p);                maxp = max(maxp, p);                if(maxp - minp == j - i)                    ans++;            }        }        printf("%d\n", ans);    }    return 0;}

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