13. 分治

分治，字⾯上的解释是「分⽽治之」，就是把⼀个复杂的问题分成两个或更多的相同的⼦问题，直到 最后⼦问题可以简单的直接求解，原问题的解即⼦问题的解的合并。

1.P1908 逆序对 - 洛谷

 

#include<iostream>

using namespace std;

typedef long long LL;
LL ret;

const int N = 5e5 + 10;
int n;
int a[N];
int tmp[N];

LL merge(int left, int right)
{
	//分到最后不能再继续分了
	if (left >= right)return 0;
	LL ret = 0;
	int mid = (left + right) / 2;
	ret += merge(left, mid);
	ret += merge(mid + 1, right);

	int cur1 = left, cur2 = mid + 1, i = left;

	while (cur1 <= mid && cur2 <= right)
	{
		if (a[cur1] <= a[cur2])
		{
			tmp[i++] = a[cur1++];

		}
		else
		{
			tmp[i++] = a[cur2++];
			ret += mid - cur1 + 1;
		}
	}
	while (cur1 <= mid)tmp[i++] = a[cur1++];
	while (cur2 <= right)tmp[i++] = a[cur2++];

	for (int i = left; i <= right; i++)
	{
		a[i] = tmp[i];
	}
	return ret;
}


int main()
{
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	ret = merge(1, n);
	cout << ret << endl;
}

2.P1923 【深基9.例4】求第 k 小的数 - 洛谷

#define _CRT_SECURE_NO_WARNINGS
#include<iostream>
#include<ctime>
using namespace std;

int n,k;
const int N = 5e6 + 10;
int a[N];

int get_random(int l, int r)
{
	return a[rand() % (r - l + 1) + l];
}

int quick_sort(int left, int right, int k)
{
	//if (left == right)return a[left];
	int p = get_random(left,right);
	int l = left - 1, r = right + 1, i = left;
	while (i < r)
	{
		if (a[i] < p)
		{
			swap(a[++l], a[i++]);
		}
		else if (a[i] == p)
		{
			i++;
		}
		else
		{
			swap(a[--r], a[i]);
		}
	}
	//这个时候就区分成3个区域
	//l - left + 1,r - l - 1;right - r +1
	int a1 = l - left + 1, b1 = r - l - 1, c1 = right - r + 1;
	if (k <= a1)return quick_sort(left, l, k);//记得写上返回
	else if (k <= a1 + b1)
	{
		return  p;
	}
	else
	{
		return	quick_sort(r, right, k - a1 - b1);//记得写上返回
	}
}

int main()
{
	srand(time(0));
	scanf("%d", &n);
	scanf("%d", &k);
	k++;//由于题意，这里的k要++
	for (int i = 1; i <= n; i++)
	{
		scanf("%d", &a[i]);
	}
	printf("%d", quick_sort(1, n, k));
}

 3.P1115 最大子段和 - 洛谷

 

#include<iostream>

using namespace std;

typedef long long LL;
int n;
const int N = 2e5 + 10;
int a[N];

LL dfs(int left,int right)
{
	if (left == right)return a[left];
	LL ret;
	int mid = (left + right) / 2;
	ret = max(dfs(left, mid), dfs(mid + 1, right));

	LL lmax = a[mid], lsum = a[mid];
	for (int i = mid - 1; i >= left; i--)
	{
		lsum += a[i];
		lmax = max(lmax, lsum);
	}
	LL rmax = a[mid + 1], rsum = a[mid + 1];
	for (int i = mid + 2; i <= right; i++)
	{
		rsum += a[i];
		rmax = max(rsum, rmax);
	}
	return max(lmax + rmax, ret);
}


int main()
{
	cin >> n;
	for (int i = 1; i <= n; i++)
	{
		cin >> a[i];
	}
	cout<<dfs(1,n);
	return 0;
}

4.P1228 地毯填补问题 - 洛谷

 

 

#include<iostream>

using namespace std;

int k,x,y;

void dfs(int a, int b, int len, int x, int y)
{
	if (len == 1)return;//除去的条件
	len = len / 2;
	//左上角
	if (x < a + len && y < b + len)
	{
		//出对角，
		cout << a + len << " " << b + len << " " << 1 << endl;
		dfs(a, b, len, x, y);//光处理其他三个角了，忘记处理所属区域的角了
		dfs(a, b + len, len, a + len - 1, b + len);
		dfs(a + len, b, len, a + len, b + len - 1);
		dfs(a + len, b + len, len, a + len, b + len);
	}
	else if (x >= a + len && y >= b + len)
	{
		cout << a + len - 1 << " " << b + len - 1 << " " << 4 << endl;
		dfs(a + len, b + len, len, x, y);
		dfs(a, b, len, a + len - 1, b + len - 1);
		dfs(a, b + len, len, a + len - 1, b + len);
		dfs(a + len, b, len, a + len, b + len - 1);
	}
	else if (x >= a + len)
	{
		cout << a + len - 1 << " " << b + len << " " << 3 << endl;
		dfs(a + len, b, len, x, y);
		dfs(a, b, len, a + len - 1, b + len - 1);
		dfs(a, b + len, len, a + len - 1, b + len);
		dfs(a + len, b + len, len, a + len, b + len);
	}
	else
	{
		cout << a + len << " " << b + len - 1 << " " << 2 << endl;
		dfs(a, b + len, len, x, y);
		dfs(a, b, len, a + len - 1, b + len - 1);
		dfs(a + len, b, len,a + len, b + len - 1);
		dfs(a + len, b + len, len, a + len, b + len);
	}
}


int main()
{
	cin >> k >> x >> y;
	k = (1 << k);
	dfs(1, 1, k, x, y);
}