poj1007【求逆序数】

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：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 105;
char str[maxn][55];
int num[300];

int n, m;
int ans[10];
int ord(int k) {
    int cnt = 0;
    for(int i = 0; i < n; i++) {
        int x = num[(int)str[k][i]];
        for(int j = 4; j > x; j--) {
            cnt += ans[j];
        }
        ans[x]++;
    }
    return cnt;
}
void init() {
    memset(ans, 0, sizeof(ans));
    num[(int)'A'] = 1;
    num[(int)'C'] = 2;
    num[(int)'G'] = 3;
    num[(int)'T'] = 4;
}

struct Node {
    int id;
    int num1;
}node[maxn];

bool cmp(Node n1, Node n2) {
    if(n1.num1 != n2.num1) return n1.num1 < n2.num1;
    return n1.id < n2.id;
}


int main() {
    init();
    while(EOF != scanf("%d %d",&n, &m) ) {
        for(int i = 0; i < m; i++) {
            scanf("\n%s", str[i]);
        }
        for(int i = 0; i < m; i++) {
            memset(ans, 0, sizeof(ans));
            node[i].id = i;
            node[i].num1 = ord(i);
        }
        sort(node, node + m, cmp);
        for(int i = 0; i < m;i++) {
            printf("%s\n", str[node[i].id]);
        }
    }
    return 0;
}

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int lowbit(int x) {
    return x & ( - x );
}

int c[305];
void Add(int x, int val) {
    while(x > 0) {
        c[x] += val;
        x -= lowbit(x);
    }
}

int Sum(int x) {
    int sum = 0;
    while(x <= 300) {
        sum += c[x];
        x += lowbit(x);
    }
    return sum;
}    


char str[105][55];
struct Node {
    int id;
    int ord;
}node[105];

bool cmp(Node n1, Node n2) {
    if(n1.ord != n2.ord) return n1.ord < n2.ord;
    return n1.id < n2.id;
}

int main() {
    int n, m;
    while(EOF != scanf("%d %d",&n, &m) ) {
        for(int i = 0; i < m; i++) {
            scanf("\n%s", str[i]);
            memset(c, 0, sizeof(c));
            int sum = 0;
            for(int j = 0; j < n; j++) {
                sum += Sum(str[i][j]);
                sum -= c[str[i][j]];
                Add(str[i][j], 1);
            }
            node[i].id = i;
            node[i].ord = sum;
        }
        sort(node, node + m, cmp);
        for(int i = 0; i < m; i++) {
            printf("%s\n", str[node[i].id]);
        }
    }
    return 0;
}