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 6 AACATGAAGG TTTTGGCCAA TTTGGCCAAA GATCAGATTT CCCGGGGGGA ATCGATGCAT
输出示例
CCCGGGGGGA AACATGAAGG GATCAGATTT ATCGATGCAT TTTTGGCCAA TTTGGCCAAA
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 <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <cstring>
#include <string>
#include <map>
#include <set>
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 60
int n, l;
char S[maxn][maxl];
vector <int> 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;
}