UVa 1481 Genome Evolution 解题报告（枚举）

1481 - Genome Evolution

Time limit: 3.000 seconds

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(2n3000). 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： 两个1-n的排列，问元素相同的长度大于等于2的连续子序列有多少对。

    做法其实很简单，别想太复杂。首先记录排列b中1-n各个数字出现的位置。在排列a中枚举起点，找出相应的在b中的位置。随着排列a元素的增加，更新对应的排列b中的最小位置和最大位置。如果二者长度相等，说明两个子排列元素相等。代码如下：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <cmath>#include <vector>#include <queue>#include <map>#include <set>#include <string>using namespace std;#define ff(i, n) for(int i=0;i<(n);i++)#define fff(i, n, m) for(int i=(n);i<=(m);i++)#define dff(i, n, m) for(int i=(n);i>=(m);i--)typedef long long LL;typedef unsigned long long ULL;void work();int main(){#ifdef ACM    freopen("in.txt", "r", stdin);//    freopen("in.txt", "w", stdout);#endif // ACM    work();}/*****************************************/int a[3333], b[3333];int pos[3333];void work(){    int n;    while(~scanf("%d", &n) && n)    {        fff(i, 1, n) scanf("%d", a+i);        fff(i, 1, n) scanf("%d", b+i), pos[b[i]] = i;        int ans = 0;        fff(i, 1, n)        {            int minP = pos[a[i]];            int maxP = pos[a[i]];            fff(j, i+1, n)            {                int p = pos[a[j]];                minP = min(minP, p);                maxP = max(maxP, p);                if(maxP - minP == j - i)                    ans++;            }        }        printf("%d\n", ans);    }}