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;
}
原文:http://www.cnblogs.com/Running-Time/p/4803593.html