【poj1007】 DNA Sorting

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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;}

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