逆序数 HDOJ 4911 Inversion

 

题目传送门

题意：可以交换两个相邻的数字顺序k次，问最后逆序对最少有多少

分析：根据逆序数的定理如果逆序数大于0，那么必定存在1<=i<n使得i和i+1交换后逆序数减1假设原逆序数为cnt，这样的话，我们就可以得到答案是max（cnt-k,0）求逆序数可以用归并的方法

 

代码：

/************************************************
* Author        :Running_Time
* Created Time  :2015/9/12 星期六 20:31:07
* File Name     :J.cpp
 ************************************************/

#include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std;

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int a[N], L[N/2], R[N/2];
ll cnt;

void Merge(int *a, int p, int q, int r)   {
    int n1 = q - p + 1, n2 = r - q;
    for (int i=1; i<=n1; ++i)   L[i] = a[p+i-1];
    for (int i=1; i<=n2; ++i)   R[i] = a[q+i];
    L[n1+1] = R[n2+1] = INF;
    for (int i=1, j=1, k=p; k<=r; ++k)  {
        if (L[i] <= R[j])    a[k] = L[i++];
        else    {
            a[k] = R[j++];  cnt += n1 - i + 1;
        }
    }
}

void Merge_Sort(int *a, int p, int r)   {
    if (p < r)  {
        int q = (p + r) >> 1;
        Merge_Sort (a, p, q);
        Merge_Sort (a, q+1, r);
        Merge (a, p, q, r);
    }
}

int main(void)    {
    int n;  ll k;
    while (scanf ("%d%I64d", &n, &k) == 2) {
        for (int i=1; i<=n; ++i)    scanf ("%d", &a[i]);
        cnt = 0;
        Merge_Sort (a, 1, n);
        printf ("%I64d\n", max (cnt - k, (ll) 0));
    }
    
    return 0;
}