UVa 1481 Genome Evolution 解题报告（枚举）

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生物学家Xi在简化视角下研究染色体之间的进化距离，通过计算两个染色体中相同基因序列的连续子序列对数，探讨相似性度量。此算法在生物学与遗传学领域具有重要意义。

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 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 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

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);
    }
}