题意：
给出一个长为n的序列，每有一个逆序对需要付出x代价，交换相邻两个数需要y代价，问最小代价。
思路：
注意到逆序对数等于交换次数，所以求出逆序对数*min(x,y)即是答案。逆序对可以通过离散化+树状数组或者归并排序求出。
代码：
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <cstring>
#include <cctype>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <map>
#include <list>
#include <set>
#include <stack>
#include <queue>
#include <string>
#include <sstream>
#define pb push_back
#define X first
#define Y second
#define lch (o<<1)
#define rch (o<<1|1)
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define pii pair<int,int>
#define qclear(a) while(!a.empty())a.pop();
#define lowbit(x) (x&-x)
#define sd(n) scanf("%d",&n)
#define sdd(n,m) scanf("%d%d",&n,&m)
#define sddd(n,m,k) scanf("%d%d%d",&n,&m,&k)
#define mst(a,b) memset(a,b,sizeof(a))
#define cout3(x,y,z) cout<<x<<" "<<y<<" "<<z<<endl
#define cout2(x,y) cout<<x<<" "<<y<<endl
#define cout1(x) cout<<x<<endl
#define IOS std::ios::sync_with_stdio(false)
#define SRAND srand((unsigned int)(time(0)))
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
using namespace std;
const double PI=acos(-1.0);
const int INF=0x3f3f3f3f;
const ll mod=1e9+7;
const double eps=1e-8;
const int maxn=100005;
const int maxm=10005;
struct node{
int val,pos;
bool operator < (const node& o)const{
return val<o.val;
}
};
int n,x,y;
node a[maxn];
int b[maxn];
int bit[maxn];
void add(int pos,int val){
while(pos<=n){
bit[pos]+=val;
pos+=lowbit(pos);
}
}
int sum(int pos){
int res=0;
while(pos){
res+=bit[pos];
pos-=lowbit(pos);
}
return res;
}
void solve() {
while(~sddd(n,x,y)){
for(int i=1;i<=n;i++){
sd(a[i].val);
a[i].pos=i;
}
sort(a+1,a+n+1);
int now=1;
b[a[1].pos]=1;
for(int i=2;i<=n;i++){
if(a[i].val!=a[i-1].val)now++;
b[a[i].pos]=now;
}
mst(bit,0);
ll ans=0;
for(int i=1;i<=n;i++){
add(b[i],1);
ans+=(i-sum(b[i]));
}
ans=ans*min(x,y);
printf("%lld\n",ans);
}
return ;
}
int main() {
#ifdef LOCAL
freopen("in.txt","r",stdin);
// freopen("out.txt","w",stdout);
#else
// freopen("","r",stdin);
// freopen("","w",stdout);
#endif
solve();
return 0;
}