【poj1007】 DNA Sorting

http://poj.org/problem?id=1007 (题目链接)

题意

　　给出m个字符串，将其按照逆序对个数递增输出。

Solution

　　树状数组经典应用。

代码

// poj1007
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<string>
#include<map>
#define MOD 1000000007
#define inf 2147483640
#define LL long long
#define free(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout);
using namespace std;
inline LL getint() {
    LL x=0,f=1;char ch=getchar();
    while (ch>'9' || ch<'0') {if (ch=='-') f=-1;ch=getchar();}
    while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

const int maxn=200;
struct data {int cnt;char s[110];}a[maxn];
int n,m,c[10];

bool cmp(data a,data b) {return a.cnt<b.cnt;}
int lowbit(int x) {return x&-x;}
void add(int x) {
    for (int i=x;i<=4;i+=lowbit(i)) c[i]++;
}
int query(int x) {
    int s=0;
    for (int i=x;i>=1;i-=lowbit(i)) s+=c[i];
    return s;
}
int main() {
    scanf("%d%d",&n,&m);
    for (int i=1;i<=m;i++) scanf("%s",a[i].s);
    for (int i=1;i<=m;i++) {
        for (int j=0;j<=5;j++) c[j]=0;
        for (int j=n-1;j>=0;j--) {
            int x;
            if (a[i].s[j]=='A') x=1;
            else if (a[i].s[j]=='C') x=2;
            else if (a[i].s[j]=='G') x=3;
            else x=4;
            add(x);
            if (x>1) a[i].cnt+=query(x-1);
        }
    }
    sort(a+1,a+1+m,cmp);
    for (int i=1;i<=m;i++) printf("%s\n",a[i].s);
    return 0;
}