逆序数

使用归并排序实现。

线段树还不会。

POJ 1804 裸题

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

int num[1001];
int count;

void merge(int left, int mid, int right)
{
	int temp[1001];
	int i = left, j = mid + 1, k = left;
	while (i <= mid&&j <= right)
	{
		if (num[i] > num[j])
		{
			count = count + mid - i + 1;
			temp[k++] = num[j++];
		}
		else
			temp[k++] = num[i++];
	}

	while (i <= mid) temp[k++] = num[i++];
	while (j <= right) temp[k++] = num[j++];

	for (i = left; i <= right; i++)
		num[i] = temp[i];
}

void mergeSort(int left, int right)
{
	if (left < right)
	{
		int mid = (left + right) / 2;
		mergeSort(left, mid);
		mergeSort(mid + 1, right);
		merge(left, mid, right);
	}
}

int main()
{
	int n, i;
	scanf("%d", &n);
	int m, j;
	for (i = 1; i <= n; i++)
	{
		scanf("%d", &m);
		for (j = 1; j <= m; j++)
			scanf("%d", &num[j]);

		count = 0;
		mergeSort(1, m);

		printf("Scenario #%d:\n", i);
		printf("%d\n\n", count);
	}
	return 0;
}

上交OJ 1219 变形题

用到了c++的二分函数。感觉STL还是强。但是有的学校不允许使用。这道题调了很长时间。

注意最后的结果用int会溢出，导致错误。

#include <stdio.h>
#include <stdlib.h>
#include <algorithm>

using namespace std;

int num[200001];
long long cnt;

void merge(int left, int mid, int right)
{
	int temp[200001];
	int i = left, j = mid + 1, k = left;
	while (i <= mid&&j <= right)
	{
		if (num[i] > num[j])
		{	
			/* tle
			int t;
			for (t = i; t <= mid; t++)
			{
				if (num[t] > (2 * num[j]))
					cnt++;
			}*/
			
			// 下面两行代码是关键
			int pos = upper_bound(num + i, num + (mid + 1), 2 * num[j]) - num;
			cnt += mid - pos + 1;

			temp[k++] = num[j++];
		}
		else
			temp[k++] = num[i++];
	}

	while (i <= mid) temp[k++] = num[i++];
	while (j <= right) temp[k++] = num[j++];

	for (i = left; i <= right; i++)
		num[i] = temp[i];
}

void mergeSort(int left, int right)
{
	if (left < right)
	{
		int mid = (left + right) / 2;
		mergeSort(left, mid);
		mergeSort(mid + 1, right);
		merge(left, mid, right);
	}
}

int main()
{	
	int m, j;
	while (EOF != scanf("%d", &m))
	{
		for (j = 1; j <= m; j++)
			scanf("%d", &num[j]);

		cnt = 0;
		mergeSort(1, m);

		printf("%lld\n", cnt);
	}

	return 0;
}