[POJ1007]DNA Sorting-白红宇

[POJ1007]DNA Sorting

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发布时间：2019-06-07

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[POJ1007]DNA Sorting

试题描述

One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC'', this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG'' has only one inversion (E and D)---it is nearly sorted---while the sequence ``ZWQM'' has 6 inversions (it is as unsorted as can be---exactly the reverse of sorted).

You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness'', from ``most sorted'' to ``least sorted''. All the strings are of the same length.

输入

The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (0 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.

输出

Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. Since two strings can be equally sorted, then output them according to the orginal order.

输入示例

10 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

输出示例

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

数据规模及约定

见“输入”

题解？

做这题练英语玩玩。并不知道这题跟 DNA 有毛关系。。。

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            
             #include 
             
              #include 
              #include 
               
                using namespace std;const int BufferSize = 1 << 16;char buffer[BufferSize], *Head, *Tail;inline char Getchar() { if(Head == Tail) { int l = fread(buffer, 1, BufferSize, stdin); Tail = (Head = buffer) + l; } return *Head++;}int read() { int x = 0, f = 1; char c = getchar(); while(!isdigit(c)){ if(c == '-') f = -1; c = getchar(); } while(isdigit(c)){ x = x * 10 + c - '0'; c = getchar(); } return x * f;}#define maxn 110#define maxl 60int n, l;char S[maxn][maxl];vector 
                
                  id[maxn*maxn];void process(int x) { int A = 0, C = 0, G = 0, sum = 0; for(int i = l; i >= 1; i--) { if(S[x][i] == 'A') A++; if(S[x][i] == 'C') sum += A, C++; if(S[x][i] == 'G') sum += A + C, G++; if(S[x][i] == 'T') sum += A + C + G; } id[sum].push_back(x); return ;}int main() { l = read(); n = read(); for(int i = 1; i <= n; i++) scanf("%s", S[i] + 1), process(i); for(int i = 0; i <= l * l; i++) for(int j = 0; j < id[i].size(); j++) printf("%s\n", S[id[i][j]] + 1); return 0;}

转载于:https://www.cnblogs.com/xiao-ju-ruo-xjr/p/5813812.html

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