codechef Little Elephant and Permutations题解

The Little Elephant likes permutations. This time he has a permutation A[1], A[2], ..., A
of numbers 1, 2, ...,N.

He calls a permutation A good, if the number of its inversions is equal to the number of its local inversions. The number of inversions is equal to the number of pairs of integers (i; j) such that 1 ≤ i < j ≤ N and A[i] > A[j],
and the number of local inversions is the number of integers i such that 1 ≤ i < N and A[i] > A[i+1].

The Little Elephant has several such permutations. Help him to find for each permutation whether it is good or not. Print YES for a corresponding test case if it is good and NO otherwise.

Input

The first line of the input contains a single integer T, the number of test cases. T test cases follow. The first line of each test case contains a single integer N, the size of a permutation. The next line
contains N space separated integers A[1], A[2], ..., A
.

Output

For each test case output a single line containing the answer for the corresponding test case. It should beYES if the corresponding permutation is good and NO otherwise.

Constraints

1 ≤ T ≤ 474

1 ≤ N ≤ 100

It is guaranteed that the sequence A[1], A[2], ..., A
is a permutation of numbers 1, 2, ..., N.

Example

Input:
4
1
1
2
2 1
3
3 2 1
4
1 3 2 4

Output:
YES
YES
NO
YES

总结一下有下面关系：

全局逆序数 == 起泡排序交换数据的交换次数

本题数据量小，能够使用暴力法计算。

可是这样时间效率是O(N*N)，要想减少到O(nlgn)那么就要使用merger sort。

以下一个类中暴力法和merge sort方法都包含了。

#pragma once
#include <stdio.h>
#include <stdlib.h>

class LittleElephantAndPermutations
{
int localInverse(const int *A, const int n)
{
int ans = 0;
for (int i = 1; i < n; i++)
{
if (A[i-1] > A[i]) ans++;
}
return ans;
}

int globalInverse(const int *A, const int n)
{
int ans = 0;
for (int i = 0; i < n; i++)
{
for (int j = i+1; j < n; j++)
{
if (A[i] > A[j]) ans++;
}
}
return ans;
}

int *mergeArr(int *left, int L, int *right, int R, int &c)
{
int *lr = new int[L+R];
int i = 0, j = 0, k = 0;
while (i < L && j < R)
{
if (left[i] <= right[j])
{
lr[k++] = left[i++];
}
else
{
lr[k++] = right[j++];
c += L-i;
}
}
while (i < L)
{
lr[k++] = left[i++];
}
while (j < R)
{
lr[k++] = right[j++];
}
return lr;
}

int biGlobalInverse(int *&A, const int n)
{
if (n <= 1) return 0;

int mid = (n-1) >> 1;//记得n-1
int *left = new int
;
int *right = new int
;
int L = 0, R = 0;

for ( ; L <= mid; L++)
{
left[L] = A[L];
}
for (int i = mid+1; i < n; i++, R++)
{
right[R] = A[i];
}
delete [] A;

int c1 = biGlobalInverse(left, L);
int c2 = biGlobalInverse(right, R);

int c = 0;
A = mergeArr(left, L, right, R, c);

delete left;
delete right;

return c1+c2+c;
}

public:
void run()
{
int T = 0, N = 0;
scanf("%d", &T);
while (T--)
{
scanf("%d", &N);
int *A = new int
;

for (int i = 0; i < N; i++)
{
scanf("%d", &A[i]);
}
int loc = localInverse(A, N);
int glo = biGlobalInverse(A, N);
if (loc == glo) puts("YES");
else puts("NO");
}
}
};

int littleElephantAndPermutations()
{
LittleElephantAndPermutations le;
le.run();
return 0;
}