UVA 11039-Building designing【贪心+绝对值排序】

UVA11039-Building designing

Time limit: 3.000 seconds

An architect wants to design a very high building. The building will consist of some ﬂoors, and each ﬂoor has a certain size. The size of a ﬂoor must be greater than the size of the ﬂoor immediately above it. In addition, the designer (who is a fan of a famous Spanish football team) wants to paint the building in blue and red, each ﬂoor a colour, and in such a way that the colours of two consecutive ﬂoors are diﬀerent. To design the building the architect has n available ﬂoors, with their associated sizes and colours. All the available ﬂoors are of diﬀerent sizes. The architect wants to design the highest possible building with these restrictions, using the available ﬂoors.
Input
The input ﬁle consists of a ﬁrst line with the number p of cases to solve. The ﬁrst line of each case contains the number of available ﬂoors. Then, the size and colour of each ﬂoor appear in one line. Each ﬂoor is represented with an integer between -999999 and 999999. There is no ﬂoor with size 0. Negative numbers represent red ﬂoors and positive numbers blue ﬂoors. The size of the ﬂoor is the absolute value of the number. There are not two ﬂoors with the same size. The maximum number of ﬂoors for a problem is 500000.
Output
For each case the output will consist of a line with the number of ﬂoors of the highest building with the mentioned conditions.
Sample Input
2

-2

-3

-9

-15

4
Sample Output
2

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1980

题意：建一栋楼，负数代表一种颜色，正数代表另一种颜色，要正负号交替且绝对值递增。

两种解法：

第一种就是简单排下序，然后贪心的思想不断找绝对值最小的就可以了，注意的就是先取正值和先取负值的情况都要考虑

还有一种就是直接对绝对值进行排序操作，然后标记正负号(貌似更简练，感谢欧尼酱的点播)

解法1AC代码：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 inline int read()
 5 {
 6     int x=0,f=1;
 7     char ch=getchar();
 8     while(ch<'0'||ch>'9')
 9     {
10         if(ch=='-')
11             f=-1;
12         ch=getchar();
13     }
14     while(ch>='0'&&ch<='9')
15     {
16         x=x*10+ch-'0';
17         ch=getchar();
18     }
19     return x*f;
20 }
21 inline void write(int x)
22 {
23     if(x<0)
24     {
25         putchar('-');
26         x=-x;
27     }
28     if(x>9)
29     {
30         write(x/10);
31     }
32     putchar(x%10+'0');
33 }
34 const int N=555555;
35 int a[N],b[N];
36 int main()
37 {
38     int t,n,p1,p2,num;
39     t=read();
40     while(t--)
41     {
42         n=read();
43         p1=0;
44         p2=0;
45         for(int i=1;i<=n;i++)
46         {
47             scanf("%d",&num);
48             if(num>0)
49                 a[p1++]=num;
50             else
51                 b[p2++]=-num;
52         }
53         sort(a,a+p1);
54         sort(b,b+p2);
55         int m1=0,m2=0,n1=1,n2=1;
56         while(m1<p1&&m2<p2)
57         {
58             while(m2<p2&&b[m2]<=a[m1])
59                 m2++;
60             if(m2!=p2)
61                 n1++;
62             else
63                 break;
64             while(m1<p1&&a[m1]<=b[m2])
65                 m1++;
66             if(m1!=p1)
67                 n1++;
68             else
69                 break;
70         }
71         m1=0,m2=0;
72         while(m1<p1&&m2<p2)
73         {
74             while(m1<p1&&a[m1]<=b[m2])
75                 m1++;
76             if(m1!=p1)
77                 n2++;
78             else
79                 break;
80             while(m2<p2&&b[m2]<=a[m1])
81                 m2++;
82             if(m2!=p2)
83                 n2++;
84             else
85                 break;
86         }
87         printf("%d\n",max(n1,n2));
88     }
89     return 0;
90 }

解法2AC代码：(欧尼酱的代码，参考一下)

 1 #include<bits/stdc++.h>
 2 const int N=5*1e5+10;
 3 using namespace std;
 4 int a[N];
 5 bool cmp(int a,int b)
 6 {
 7     return abs(a)<abs(b);
 8 }
 9 int main()
10 {
11     int t,n,flag,num;
12     while(~scanf("%d",&t))
13     {
14         while(t--)
15         {
16             scanf("%d",&n);
17             flag=0;
18             num=0;
19             for(int i=0;i<n;i++)
20                 scanf("%d",&a[i]);
21             sort(a,a+n,cmp);
22             if(a[0]>0)
23                 flag=1;
24             else 
25                 flag=2;
26             num++;
27             for(int i=1;i<n;i++)
28             {
29                 if(flag==1)
30                 {
31                     if(a[i]<0)
32                     {
33                         flag=2;
34                         num++;
35                     }
36                 }
37                 else if(flag==2)
38                 {
39                     if(a[i]>0)
40                     {
41                         flag=1;
42                         num++;
43                     }
44                 }
45              }
46           printf("%d\n",num);
47         }
48     }
49     return 0;
50 }