HOJ 2430——Counting the algorithms(树状数组+贪心)
2014-08-28 19:00
381 查看
Counting the algorithms
My Tags | (Edit) |
---|
Source : mostleg | |||
Time limit : 1 sec | Memory limit : 64 M |
As most of the ACMers, wy's next target is algorithms, too. wy is clever, so he can learn most of the algorithms quickly. After a short time, he has learned a lot. One day, mostleg asked him that how many he had
learned. That was really a hard problem, so wy wanted to change to count other things to distract mostleg's attention. The following problem will tell you what wy counted.
Given 2N integers in a line, in which each integer in the range from 1 to N will appear exactly twice. You job is to choose
one integer each time and erase the two of its appearances and get a mark calculated by the differece of there position. For example, if the first 3 is in position 86 and
the second 3 is in position 88, you can get 2 marks if you choose to erase 3 at this time. You should notice that after one
turn of erasing, integers' positions may change, that is, vacant positions (without integer) in front of non-vacant positions is not allowed.
Input
There are multiply test cases. Each test case contains two lines.
The first line: one integer N(1 <= N <= 100000).
The second line: 2N integers. You can assume that each integer in [1,N] will appear just twice.
Output
One line for each test case, the maximum mark you can get.
Sample Input
3 1 2 3 1 2 3 3 1 2 3 3 2 1
Sample Output
6 9
Hint
We can explain the second sample as this. First, erase 1, you get 6-1=5 marks. Then erase 2, you get 4-1=3 marks.
You may notice that in the beginning, the two 2s are at positions 2 and 5, but at this time, they are at positions 1 and 4.
At last erase 3, you get 2-1=1marks. Therefore, in total you get 5+3+1=9 and that is the best strategy.
————————————————————分割线——————————————————
题目大意:
给定2n个整数,范围为1到n,每个数出现两次,每个数都有一个下标,每次删除两个相同的数,能获得下标差的值,求这个值得最大
思路:
对于相间的情况,从外围往内删跟从内往外删是一样的,而对于 包含的情况,从外往内删是最优的。因此,选择从外往内删。
具体操作:
扫描一次,记录每个数第二次出现的下标,将数的下标插人树状数组。然后从左往右或者从右往左扫描,那么值为两个数的距离差,之后删除其中一个数,如果是从左往右扫描的,则删除第二次出现的下标,如果是从右往左扫描的,则删除第一次出现的下标
#include<iostream> #include<cstring> #include<cstdio> #define ll long long #define M 200000+10 using namespace std; int a[M],c[M],f[M]; bool judge[M]; int n; void update(int x,int v) { for(int i=x;i<=2*n;i+=i&-i){ c[i]+=v; } } int getsum(int x) { int sum=0; for(int i=x;i>0;i-=i&-i){ sum+=c[i]; } return sum; } int main() { while(scanf("%d",&n)!=EOF){ memset(judge,false,sizeof(judge)); memset(c,0,sizeof(c)); memset(f,0,sizeof(f)); for(int i=1;i<=2*n;++i){ scanf("%d",&a[i]); update(i,1); if(!judge[a[i]]) judge[a[i]]=true; else f[a[i]]=i; } int sum=0; for(int i=1;i<=2*n;++i){ sum+=getsum(f[a[i]])-getsum(i); update(f[a[i]],-1); } printf("%d\n",sum); } return 0; }
相关文章推荐
- hoj 2430 Counting the algorithms(树状数组,贪心)
- hoj 2430 Counting the algorithms
- hoj 2430 Counting the algorithms
- HOJ——T 2430 Counting the algorithms
- HOJ_2430 Counting the algorithms
- hoj 2430 Counting the algorithms (树状数组)
- 【HOJ2430】【贪心+树状数组】 Counting the algorithms
- hoj2430 Counting the algorithms
- 【HOJ2430】Counting the Algorithms-区间维护
- HOJ2430 Counting the Algorithms 树状数组
- JAVA-勇者斗恶龙-The Dragon of Loowater,Uva 11292-排序后贪心
- hdoj 1052Tian Ji -- The Horse Racing 【贪心】
- HDU 1052 - Tian Ji -- The Horse Racing(贪心)
- GYM 101102 B.The Little Match Girl (贪心)
- HDOJ 1052 Tian Ji -- The Horse Racing 贪心
- Hoj 3136 To help the beauty
- POJ 2282 The Counting Problem
- CodeForces 349B--Color the Fence(贪心)
- POJ-2057 The Lost House 贪心在动态规划中的应用
- codeforces349B - Color the Fence 贪心+DP