POJ—DNA Sorting
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DNA SortingTime Limit: 1000MS Memory Limit: 10000KTotal Submissions: 103622 Accepted: 41507
DescriptionOne 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.
Input
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.
Output
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.
Sample Input
10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT
Sample Output
CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA
终于做完的题目,长吁一口气。
1.通过这道题再学了map,懂了基本用法,终于完全懂了基本用法了(是学长教的,哈哈哈)
以前总觉得像隔着雾气氤氲的玻璃看,现在终倒是敲破了玻璃。
希望能以后能敲破越来越多的玻璃。
2.get到的新知识:vector<int>vec能保证没有相同的数据。
map<int,string> mp;WA//正因为可能出现相同的测试案例
map<vector<int>,string>//不会出现相同的测试案例,AC
附上代码:
#include<iostream>#include<cstdio>#include<string>#include<cstring>#include<map>#include<functional>#include<vector>using namespace std;int main(){ int strn,strl; string str[110]; int sortans[110]; scanf("%d%d",&strl,&strn); map<vector<int>,string> mp; for(int i=0;i<strn;i++) { int cnt=0; cin>>str[i]; vector<int>vec; for(int j=0; j<strl; j++) { for (int k = j; k<strl; k++) { if (str[i][j]>str[i][k]) cnt++; } } vec.push_back(cnt); vec.push_back(i);//判断是否有重复数据。 mp[vec]=str[i]; } for(map<vector<int>,string>::iterator it=mp.begin();it!=mp.end();it++) cout<<it->second<<endl; return 0;}
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