2018 Multi-University Training Contest 2-1010-Swaps and Inversions

题目链接http://acm.hdu.edu.cn/contests/contest_showproblem.php?pid=1010&cid=803

 

题目大意

一个乱序的序列，每个逆序对需花费x，可以花费y交换相邻数字，求最小花费

 

简单思路

找出逆序对数目m，然后比较x、y的大小，m*min(x,y)

 

ac代码

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3FFFFFFF
#define fo(i,a,b)    for(int i=a;i<=b;i++)
#define fo_(i,a,b)    for(int i=a;i>=b;i--)
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define M_(a) memset(a,-1,sizeof a)
#define pb push_back
const ll mod=1000000007;
ll powmod(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;}
#define maxn 100010
#define MAX_VALUE 12343564

int x,y,n;
int a[maxn];
ll cnt;

void merge(int *a , int first, int mid, int last)
{
    int * temp = new int[last - first + 1];     
    int first1 = first, last1 = mid;
    int first2 = mid + 1, last2 = last;
    int index = 0;
    while(first1 <= last1 && first2 <= last2)
    {
        if(a[first1 ] <= a[first2])
        {
            temp[index++] = a[first1++];
        }
        else
        {
            cnt += last1 - first1 + 1;
            temp[index++] = a[first2++];
        }
    }
    while(first1 <= last1)
    {
        temp[index++] = a[first1++];
    }
    while(first2 <= last2)
    {
        temp[index++] = a[first2++];
    }
    int i;
    for(i = first; i <= last; i++)
    {
        a[i] = temp[i - first];
    }
    delete [] temp;
    return;
}
void inv_pair(int *a, int first, int last)
{
    if(last - first > 0)
    {
        int mid = (last + first) / 2;
        inv_pair(a, first, mid);
        inv_pair(a, mid + 1, last);
        merge(a, first, mid, last);
    }
    return;
}


int main()
{
    while(~(scanf("%d %d %d",&n,&x,&y))){
        cnt=0;
        fo(i,0,n-1){
            scanf("%d",&a[i]);
        }
        inv_pair(a,0,n-1);
        ll min_s;
        min_s=min(cnt*x,cnt*y);
        printf("%lld\n",min_s);
    }
    return 0;
}

 

 

总结

解题关键在于找出逆序对