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
5
7
-2
6
9
-3
8
11
-9
2
5
18
17
-15
4
Sample Output
2
5
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 ",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 ",num); 47 } 48 } 49 return 0; 50 }