Frosh Week

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题解：求逆序数问题，我用树状数组和归并来做的，因为没有告诉编号范围，所以先离散化在做。例如：7,3,5,77离散化之后：3,1,2,4，他们的逆序数一样的。可以理解为将原来的数改成离散后的数。

树状数组：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define mem(a) memset(a,0,sizeof(a));

using namespace std;

struct Node
{
	int pos;
	long long value;
	bool operator< (Node t) const
	{
		return value < t.value;
	}
};

Node a[1000005];
int b[1000005];
int c[100005];

void update(int x,int n,int plus)
{
	while(x <= n)
	{
		c[x] += plus;
		x += x & (-x);
	}
}

long long getSum(int x)
{
	long long res = 0;
	while(x > 0)
	{
		res += c[x];
		x -= x & (-x);
	}
	
	return res;
}

int main()
{
	int n;
	while(scanf("%d",&n) != EOF)
	{
		for(int i = 1;i <= n;i++)
		{
			scanf("%d",&a[i].value);
			a[i].pos = i;
		}
		
		sort(a + 1,a + n + 1);
		
		for(int i = 1;i <= n;i++)
		{
			b[a[i].pos] = i;
		}
		
		mem(c);
		long long sum = 0;
		for(int i = 1;i <= n;i++)
		{
			update(b[i],n,1);
			sum += i - getSum(b[i]);
		}
		
		printf("%lld\n",sum);
	}
	
	return 0;
}

归并：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define mem(a) memset(a,0,sizeof(a));

using namespace std;

struct Node
{
	int pos;
	long long value;
	bool operator< (Node t) const
	{
		return value < t.value;
	}
};

Node a[1000005];
int b[1000005];
int c[100005];

long long res;

void merge(int l,int r)
{
	int mid = (l + r) >> 1;
	int low = l;
	int high = mid + 1;
	int k = l;
	while(low <= mid && high <= r)
	{
		if(b[low] <= b[high])
		{
			c[k++] = b[low++];
		}
		else
		{
			c[k++] = b[high++];
			res += mid - low + 1;
		}
	}
	for(int i = low;i <= mid;i++)
	{
		c[k++] = b[i];
	}
	for(int i = high;i <= r;i++)
	{
		c[k++] = b[i];
	}
	for(int i = l;i < k;i++)
	{
		b[i] = c[i];
	}
}

void mergeSort(int l,int r)
{
	if(l < r)
	{
		int mid = (l + r) >> 1;
		mergeSort(l,mid);
		mergeSort(mid + 1,r);
		merge(l,r);
	}
}

int main()
{
	int n;
	while(scanf("%d",&n) != EOF)
	{
		for(int i = 1;i <= n;i++)
		{
			scanf("%d",&a[i].value);
			a[i].pos = i;
		}
		
		sort(a + 1,a + n + 1);
		
		for(int i = 1;i <= n;i++)
		{
			b[a[i].pos] = i;
		}
		
		res = 0;
		mergeSort(1,n);
		
		printf("%lld\n",res);
	}
	
	return 0;
}

Frosh Week

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原文地址：http://blog.csdn.net/wang2534499/article/details/47377535