HDU - 6318 Swaps and Inversions（归并板子）

题目链接

找到逆序对的数量，然后乘以min(x,y);

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;

const int maxn = 100010;
ll n,x,y,ans;//ans记录逆序对的数量
ll tmp[maxn],a[maxn];

void merge(int l,int r,int m){
	int i=l,j=m+1,k=l;
	while(i<=m&&j<=r){
		if(a[i]>a[j]){
			tmp[k++]=a[j++];
			ans+=m-i+1;
		}
		else{
			tmp[k++]=a[i++];
		}
	}
	while(i<=m)tmp[k++]=a[i++];
	while(j<=r)tmp[k++]=a[j++];
	for(int i=l;i<=r;i++) a[i]=tmp[i];
}

void merges(int l,int r){
	if(l<r){
		int m=(l+r)>>1;
		merges(l,m);
		merges(m+1,r);
		merge(l,r,m);
	}
}

int main()
{
    ios::sync_with_stdio(false);
    while(cin>>n>>x>>y){
    for(int i = 0;i < n;i++)
        scanf("%lld",&a[i]);
    ans = 0;
    merges(0,n-1);
    printf("%lld\n",ans*min(x,y));

    }
    return 0;
}