BestCoder Round #65

博弈 1002 ZYB‘s Game

题意:中文

分析:假定两个人是绝顶聪明的,一定会采取最优的策略.所以如果选择X的左边的一个点,那么后手应该选择X的右边对称的点,如果没有则必输,否则必胜,然后再分析下就是奇数是1,偶数是0

树状数组+二分(逆序数) 1003 ZYB‘s Premutation

题意:已知每个点前缀逆序对数和,求原排列

分析:可以得知每个点前面有几个比它大,那么用树状数组维护,二分查询从i到n有几个数字,那么答案是i-1

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 5e4 + 5;
int A[N], B[N], a[N];
struct BIT	{
	int c[N], sz;
	void init(int n)	{
		memset (c, 0, sizeof (c));
		sz = n;
	}
	void updata(int i, int x)	{
		while (i <= sz)	{
			c[i] += x;
			i += i & (-i);
		}
	}
	int query(int i)	{
		int ret = 0;
		while (i)	{
			ret += c[i];
			i -= i & (-i);
		}
		return ret;
	}
	int bsearch(int l, int r, int k)	{
		int ret = 0, rr = r;
		while (l <= r)	{
			int mid = (l + r) >> 1;
			int cnt = query (rr) - query (mid);
			if (cnt == k)	{
				ret = mid;	r = mid - 1;
			}
			else if (cnt > k)	l = mid + 1;
			else	r = mid - 1;
		}
		return ret;
	}
}bit;
int n;

int main(void)	{
	int T;	scanf ("%d", &T);
	while (T--)	{
		scanf ("%d", &n);
		for (int i=1; i<=n; ++i)	{
			scanf ("%d", &A[i]);
		}
		B[0] = 0;
		for (int i=2; i<=n; ++i)	{
			B[i] = A[i] - A[i-1];
		}
		bit.init (n);
		for (int i=1; i<=n; ++i)	{
			bit.updata (i, 1);
		}
		for (int i=n; i>=1; --i)	{
			a[i] = bit.bsearch (1, n, B[i]);
			bit.updata (a[i], -1);
		}
		for (int i=1; i<=n; ++i)	{
			printf ("%d%c", a[i], i == n ? ‘\n‘ : ‘ ‘);
		}
	}

	return 0;
}

树形DP 1004 ZYB‘s Tree

题意:一棵树,求所有点它到其他点的距离不大于K的个数的异或和 

分析:dp[u][i] 表示u到子孙的距离为i的点的个数,dp[u][i+1] += dp[v][i].dp2[v][i] 表示v到上面的距离为i的点的个数,dp[v][i+1] += dp2[u][i] + dp[u][i] - dp[v][i-1]

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int N = 5e4 + 5;
int A[N], B[N], a[N];
struct BIT	{
	int c[N], sz;
	void init(int n)	{
		memset (c, 0, sizeof (c));
		sz = n;
	}
	void updata(int i, int x)	{
		while (i <= sz)	{
			c[i] += x;
			i += i & (-i);
		}
	}
	int query(int i)	{
		int ret = 0;
		while (i)	{
			ret += c[i];
			i -= i & (-i);
		}
		return ret;
	}
	int bsearch(int l, int r, int k)	{
		int ret = 0, rr = r;
		while (l <= r)	{
			int mid = (l + r) >> 1;
			int cnt = query (rr) - query (mid);
			if (cnt == k)	{
				ret = mid;	r = mid - 1;
			}
			else if (cnt > k)	l = mid + 1;
			else	r = mid - 1;
		}
		return ret;
	}
}bit;
int n;

int main(void)	{
	int T;	scanf ("%d", &T);
	while (T--)	{
		scanf ("%d", &n);
		for (int i=1; i<=n; ++i)	{
			scanf ("%d", &A[i]);
		}
		B[0] = 0;
		for (int i=2; i<=n; ++i)	{
			B[i] = A[i] - A[i-1];
		}
		bit.init (n);
		for (int i=1; i<=n; ++i)	{
			bit.updata (i, 1);
		}
		for (int i=n; i>=1; --i)	{
			a[i] = bit.bsearch (1, n, B[i]);
			bit.updata (a[i], -1);
		}
		for (int i=1; i<=n; ++i)	{
			printf ("%d%c", a[i], i == n ? ‘\n‘ : ‘ ‘);
		}
	}

	return 0;
}