HDU 4911 Inversion - 疯狂的癫子解题心得

原题：

Description

bobo has a sequence a ₁,a ₂,…,a _n. He is allowed to swap two adjacent numbers for no more than k times.

Find the minimum number of inversions after his swaps.

Note: The number of inversions is the number of pair (i,j) where 1≤i<j≤n and a _i>a _j.

Input

The input consists of several tests. For each tests:

The first line contains 2 integers n,k (1≤n≤10 ⁵,0≤k≤10 ⁹). The second line contains n integers a ₁,a ₂,…,a _n (0≤a_i≤10 ⁹).

Output

For each tests:

A single integer denotes the minimum number of inversions.

Sample Input

3 1 2 2 1 3 0 2 2 1

分析：

先用归并法求逆序数，再将逆序数减去k，结果为正答案即这个

结果为负答案就是0

代码：

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
__int64 a[100005], cnt, c[100005];
__int64 k;
void merg(int low, int mid, int high)
{int i = low, j = mid + 1;cnt = 0;while (i <= mid&&j <= high){if (a[i]>a[j]){c[cnt++] = a[j++];k += mid - i + 1;}else{c[cnt++] = a[i++];}}while (i <= mid){c[cnt++] = a[i++];}while (j <= high){c[cnt++] = a[j++];}cnt = 0; i = low;while (i <= high){a[i++] = c[cnt++];}
}
void merger(int low, int high)
{int mid;if (low<high){mid = (low + high) / 2;merger(low, mid);merger(mid + 1, high);merg(low, mid, high);}
}
int main()
{int i, n, m;while (~scanf("%d%d", &n, &m)){k = 0;for (i = 0; i<n; i++){scanf("%I64d", &a[i]);}merger(0, n - 1);if (k - m <= 0)printf("0\n");elseprintf("%I64d\n", k - m);}return 0;
}

转载于:https://www.cnblogs.com/shawn-ji/p/4711856.html

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