基数排序

pass

 1 #include <cstdio>
 2 #include <algorithm>
 3 using namespace std;
 4 int t1[110000],t2[110000],rk[110000],uni[10],c[10];
 5 int n;
 6 int get(int *x,int i,int o)
 7 {
 8     return x[i] / uni[o - 1] % 10;
 9 }
10 int main()
11 {
12     scanf("%d",&n);
13     for (int i = 1;i <= n;i++)
14         scanf("%d",&t1[i]);
15     uni[0] = 1;
16     for (int i = 1;i <= 9;i++) uni[i] = uni[i - 1] * 10;
17     int *x = t1,*y = t2;
18     for (int o = 1;o <= 10;o++)
19     {
20         for (int i = 0;i <= 9;i++) c[i] = 0;
21         for (int i = 1;i <= n;i++) c[get(x,i,o)]++;
22         for (int i = 0;i <= 9;i++) c[i] += c[i - 1];
23         for (int i = n;i >= 1;i--) rk[i] = c[get(x,i,o)]--;
24         for (int i = 1;i <= n;i++) y[rk[i]] = x[i];
25         swap(x,y);
26     }
27     for (int i = 1;i <= n;i++)
28         printf("%d ",t1[i]);
29     return 0;
30 }
View Code

 

#include <cstdio>
#include <algorithm>
using namespace std;
int tp[110000],sa[110000],vec[110000],uni[10],c[10];
int n;
int get(int *x,int i,int o)
{
    return x[i] / uni[o - 1] % 10;
}
int main()
{
    scanf("%d",&n);
    for (int i = 1;i <= n;i++)
        scanf("%d",&vec[i]);
    uni[0] = 1;
    for (int i = 1;i <= 9;i++) uni[i] = uni[i - 1] * 10;
    /*
    int *x = t1,*y = t2;
    for (int o = 1;o <= 10;o++)
    {
        for (int i = 0;i <= 9;i++) c[i] = 0;
        for (int i = 1;i <= n;i++) c[get(x,i,o)]++;
        for (int i = 0;i <= 9;i++) c[i] += c[i - 1];
        for (int i = n;i >= 1;i--) rk[i] = c[get(x,i,o)]--;
        for (int i = 1;i <= n;i++) y[rk[i]] = x[i];
        swap(x,y);
    }*/
    for (int i = 1;i <= n;i++) sa[i] = i;
    for (int o = 1;o <= 10;o++)
    {//tp[i]表示第二关键排名为i的第一关键字位置
        for (int i = 1;i <= n;i++) tp[i] = sa[i];
        for (int i = 0; i <= 9; i++) c[i] = 0;
        for (int i = 1; i <= n; i++) c[get(vec,i,o)]++;
        for (int i = 1; i <= 9; i++) c[i] += c[i - 1];
        for (int i = n; i >= 1; i--) sa[ c[get(vec,tp[i],o)]-- ] = tp[i];
    }
    for (int i = 1;i <= n;i++)
        printf("%d ",vec[sa[i]]);
    return 0;
}
View Code

 

posted @ 2020-02-23 16:13  IAT14  阅读(140)  评论(0)  编辑  收藏  举报