标签:style os io for amp new size type
HDU 4911 Inversion
考点:归并排序
思路:这题呀比赛的时候忘了知道可以用归并排序算出逆序数,但是忘了归并排序的实质了,然后不会做……
因为看到题上说是相邻的两个数才能交换的时候,感觉归并排序好像不是得要相邻的呀,然后就这样晕……刚才重新看了才发现,归并就是相邻的交换的,正好是用来求逆序数的,唉……真的是做这个归并排序比赛就来了……真好!
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<limits>
#include<bitset>
#define mem(a,b) memset(a,b,sizeof(a))
#define lson i<<1,l,mid
#define rson i<<1|1,mid+1,r
#define INF 510010
#define maxn 100010
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
ll sum;
void merge(ll A[], int p, int q, int r)
{
int n1 = q - p + 1;
int n2 = r - q;
ll *L=new ll[n1 + 1];
ll *R=new ll[n2 + 1];
for(int i = 0; i < n1; i++)
L[i] = A[p + i];
for(int i = 0; i < n2; i++)
R[i] = A[q + 1 + i];
L[n1] = numeric_limits<ll>::max();
R[n2] = numeric_limits<ll>::max();
int i = 0, j = 0;
for(int k = p; k <= r; k++)
{
if(L[i] <= R[j]) A[k] = L[i],i++;
else A[k] = R[j],j++,sum += n1 - i;
}
delete []L;
delete []R;
}
void mergesort(ll A[], int l, int r)
{
if(l >= r)
return ;
int q = (l + r) / 2;
mergesort(A, l, q);
mergesort(A, q + 1, r);
merge(A, l, q, r);
}
ll a[100005];
int main()
{
//freopen("test.txt","r",stdin);
int n,k,i;
while(~scanf("%d%d",&n,&k))
{
sum=0;
for(i=0; i<n; i++)
scanf("%d",a+i);
mergesort(a,0,n-1);
if(sum>k) printf("%I64d\n",sum-k);
else puts("0");
}
return 0;
}
标签:style os io for amp new size type
原文地址:http://blog.csdn.net/u011466175/article/details/38391609