poj1007【求逆序数】

时间：2014-09-26 22:07:08 阅读：194 评论：0 收藏：0 [点我收藏+]

标签：des style blog http color io os ar for

Description

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.

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

分析：求逆序数
因为数据很小只有四个字母可以直接求也可用树状数组

代码：

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 const int maxn = 105;
 8 char str[maxn][55];
 9 int num[300];
10 
11 int n, m;
12 int ans[10];
13 int ord(int k) {
14     int cnt = 0;
15     for(int i = 0; i < n; i++) {
16         int x = num[(int)str[k][i]];
17         for(int j = 4; j > x; j--) {
18             cnt += ans[j];
19         }
20         ans[x]++;
21     }
22     return cnt;
23 }
24 void init() {
25     memset(ans, 0, sizeof(ans));
26     num[(int)‘A‘] = 1;
27     num[(int)‘C‘] = 2;
28     num[(int)‘G‘] = 3;
29     num[(int)‘T‘] = 4;
30 }
31 
32 struct Node {
33     int id;
34     int num1;
35 }node[maxn];
36 
37 bool cmp(Node n1, Node n2) {
38     if(n1.num1 != n2.num1) return n1.num1 < n2.num1;
39     return n1.id < n2.id;
40 }
41 
42 
43 int main() {
44     init();
45     while(EOF != scanf("%d %d",&n, &m) ) {
46         for(int i = 0; i < m; i++) {
47             scanf("\n%s", str[i]);
48         }
49         for(int i = 0; i < m; i++) {
50             memset(ans, 0, sizeof(ans));
51             node[i].id = i;
52             node[i].num1 = ord(i);
53         }
54         sort(node, node + m, cmp);
55         for(int i = 0; i < m;i++) {
56             printf("%s\n", str[node[i].id]);
57         }
58     }
59     return 0;
60 }

View Code

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <algorithm>
 5 using namespace std;
 6 
 7 int lowbit(int x) {
 8     return x & ( - x );
 9 }
10 
11 int c[305];
12 void Add(int x, int val) {
13     while(x > 0) {
14         c[x] += val;
15         x -= lowbit(x);
16     }
17 }
18 
19 int Sum(int x) {
20     int sum = 0;
21     while(x <= 300) {
22         sum += c[x];
23         x += lowbit(x);
24     }
25     return sum;
26 }    
27 
28 
29 char str[105][55];
30 struct Node {
31     int id;
32     int ord;
33 }node[105];
34 
35 bool cmp(Node n1, Node n2) {
36     if(n1.ord != n2.ord) return n1.ord < n2.ord;
37     return n1.id < n2.id;
38 }
39 
40 int main() {
41     int n, m;
42     while(EOF != scanf("%d %d",&n, &m) ) {
43         for(int i = 0; i < m; i++) {
44             scanf("\n%s", str[i]);
45             memset(c, 0, sizeof(c));
46             int sum = 0;
47             for(int j = 0; j < n; j++) {
48                 sum += Sum(str[i][j]);
49                 sum -= c[str[i][j]];
50                 Add(str[i][j], 1);
51             }
52             node[i].id = i;
53             node[i].ord = sum;
54         }
55         sort(node, node + m, cmp);
56         for(int i = 0; i < m; i++) {
57             printf("%s\n", str[node[i].id]);
58         }
59     }
60     return 0;
61 }

View Code

poj1007【求逆序数】

标签：des style blog http color io os ar for

原文地址：http://www.cnblogs.com/zhanzhao/p/3995525.html

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