HDU 6318（树状数组）

Swaps and Inversions

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 449    Accepted Submission(s): 166

 

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

 

每次交换最多减少一个逆序对。

假设逆序对有num个，那么答案就是min(num*x,num*y)

离散化+树状数组即可

#include
#define mp make_pair
#define fir first
#define se second
#define ll long long
#define pb push_back
using namespace std;
const int maxn=5e5+10;
const ll mod=1e9+7;
const int maxm=1e6+10;
const double eps=1e-7;
const ll inf=(ll)1e13;
int n;
ll tree[maxn];
int a[maxn];
int lowbit(int x){
    return x&(-x);
}
void update(int x,int d){
    for (int i=x;i<=n;i+=lowbit(i)){
        tree[i]+=d;
    }
}
ll getsum(int x){
    ll ans=0;
    for (int i=x;i>0;i-=lowbit(i))
        ans+=tree[i];
    return ans;
}
map ma;
int b[maxn];
ll x,y;
int main(){
    while(scanf("%d %lld %lld",&n,&x,&y)!=EOF){
        for (int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            b[i]=a[i];
        }
        sort(b+1,b+n+1);
        int cnt=1;
        ma.clear();
        for (int i=1;i<=n;i++)
            ma[b[i]]=cnt++;
        for (int i=1;i<=n;i++)
            a[i]=ma[a[i]];
        for (int i=0;i<=n;i++) tree[i]=0;
        ll ans=0;
        for (int i=n;i>=1;i--){
            ans+=getsum(a[i]-1);
            update(a[i],1);
        }
        printf("%lld\n",ans*min(x,y));
    }
    return 0;
}