AcWing 788. 逆序对的数量
原题链接
简单
作者:
王一博铁粉
,
2024-02-21 11:57:11
,
所有人可见
,
阅读 30
C++ 代码
#include <iostream>
using namespace std;
typedef long long LL;
const int N = 100010;
int n;
int q[N], tmp[N];
LL merge_sort(int q[], int l, int r){
//判断该数组里面是否只有一个数
if(l>= r) return 0;
int mid = l + r >> 1;
//将逆序对全在左边和逆序对全在右边的逆序对数量加起来,存放在res里面。
LL res = merge_sort(q, l, mid) + merge_sort(q, mid + 1, r);
//归并排序,处理逆序对在左右两边的情况。核心:res = mid-i+1;
int k = 0, i = l, j = mid + 1;
while(i <= mid && j <= r){
if(q[i] <= q[j]){
tmp[k++] = q[i++];
}else{
tmp[k++] = q[j++];
res = res + (mid-i+1);
//等号右边的res指的是“逆序对全在左边和逆序对全在右边”的逆序对数量。
}
}
//扫尾
while(i<= mid){
tmp[k++] = q[i++];
}
while(j <= r){
tmp[k++] = q[j++];
}
//物归原主
for(int i = l, k = 0; i <= r; i ++){
q[i] = tmp[k++];
}
return res;
}
int main()
{
scanf("%d", &n);
for(int i = 0; i < n; i ++){
scanf("%d",&q[i]);
}
cout << merge_sort(q, 0, n-1) << endl;
return 0;
}
//错因:第24行的比较大小"if(q[i] <= q[j]){" 等于号不能省