Hdu6318Swaps and Inversions
2349 ワード
Swaps and Inversions
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
Long long ago, there was an integer sequence a. Tonyfang think this sequence is messy, so he will count the number of inversions in this sequence. Because he is angry, you will have to pay x yuan for every inversion in the sequence. You don't want to pay too much, so you can try to play some tricks before he sees this sequence. You can pay y yuan to swap any two adjacent elements. What is the minimum amount of money you need to spend? The definition of inversion in this problem is pair (i,j) which 1≤iaj.
Input
There are multiple test cases, please read till the end of input file. For each test, in the first line, three integers, n,x,y, n represents the length of the sequence. In the second line, n integers separated by spaces, representing the orginal sequence a. 1≤n,x,y≤100000, numbers in the sequence are in [−109,109]. There're 10 test cases.
Output
For every test case, a single integer representing minimum money to pay.
Sample Input
3 233 666 1 2 3 3 1 666 3 2 1
Sample Output
0 3
cntを所与の数列の逆シーケンスとする
前に見たように、隣接する2つの要素が交換される場合、数列を秩序化するにはcnt回が必要です.
では、実はans = k*(y-x) + (cnt-k)x (0<=k<=cnt)
ほら、y>xなら、交換するのに一度の費用が交換しないより高いなら、交換しないで、全部xでいいです.
そうでなければ、全部交換すれば終わりではないので、
最後の出力は簡単で、ans=min(x*cnt,y*cnt)
どのように逆順数を求めますか、ここ、
コードを少し変更すればいい
ここに私の1つを貼ります(少し違いますよ)
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 0 Accepted Submission(s): 0
Problem Description
Long long ago, there was an integer sequence a. Tonyfang think this sequence is messy, so he will count the number of inversions in this sequence. Because he is angry, you will have to pay x yuan for every inversion in the sequence. You don't want to pay too much, so you can try to play some tricks before he sees this sequence. You can pay y yuan to swap any two adjacent elements. What is the minimum amount of money you need to spend? The definition of inversion in this problem is pair (i,j) which 1≤iaj.
Input
There are multiple test cases, please read till the end of input file. For each test, in the first line, three integers, n,x,y, n represents the length of the sequence. In the second line, n integers separated by spaces, representing the orginal sequence a. 1≤n,x,y≤100000, numbers in the sequence are in [−109,109]. There're 10 test cases.
Output
For every test case, a single integer representing minimum money to pay.
Sample Input
3 233 666 1 2 3 3 1 666 3 2 1
Sample Output
0 3
cntを所与の数列の逆シーケンスとする
前に見たように、隣接する2つの要素が交換される場合、数列を秩序化するにはcnt回が必要です.
では、実はans = k*(y-x) + (cnt-k)x (0<=k<=cnt)
ほら、y>xなら、交換するのに一度の費用が交換しないより高いなら、交換しないで、全部xでいいです.
そうでなければ、全部交換すれば終わりではないので、
最後の出力は簡単で、ans=min(x*cnt,y*cnt)
どのように逆順数を求めますか、ここ、
コードを少し変更すればいい
ここに私の1つを貼ります(少し違いますよ)
#include
#include
#include
#include
#define ll long long
using namespace std;
int n,tree[100010];
void add(int k,int num)
{
while(k<=n)
{
tree[k]+=num;
k+=k&-k;
}
}
int query(int k)
{
int sum=0;
while(k)
{
sum+=tree[k];
k-=k&-k;
}
return sum;
}
struct node
{
int x,i;
bool operator < (const node &a) const
{
if(x!=a.x)
return x