8VC Venture Cup 2016 - Final Round (Div. 2 Edition)

暴力　A - Orchestra

import java.io.*;
import java.util.*;

public class Main	{
	public static void main(String[] args)	{
		Scanner cin = new Scanner (System.in);
		int r = cin.nextInt ();
		int c = cin.nextInt ();
		int n = cin.nextInt ();
		int k = cin.nextInt ();
		int[][] a = new int[12][12];
		for (int i=0; i<n; ++i)	{
			int x = cin.nextInt ();
			int y = cin.nextInt ();
			a[x][y] = 1;
		}
		int ans = 0;
		for (int sx=1; sx<=r; ++sx)	{
			for (int lx=0; sx+lx<=r; ++lx)	{
				for (int sy=1; sy<=c; ++sy)	{
					for (int ly=0; sy+ly<=c; ++ly)	{
						int cnt = 0;
						for (int i=sx; i<=sx+lx; ++i)	{
							for (int j=sy; j<=sy+ly; ++j)	{
								if (a[i][j] == 1)	cnt++;
								if (cnt >= k)	break;
							}
						}
						if (cnt >= k)	ans++;
					}
				}
			}
		}
		System.out.println (ans);
	}
}

找规律　B - Island Puzzle

相对位置不变，０的位置任意，当做不存在．查看每个点到对应的点是否能一起移动到．

#include <bits/stdc++.h>

const int N = 2e5 + 5;
int a[N], b[N];
int n;

int main()	{
	scanf ("%d", &n);
	int tot1 = 0, tot2 = 0;
	for (int x, i=0; i<n; ++i)	{
		scanf ("%d", &x);
		if (x != 0)	a[tot1++] = x;
	}
	for (int x, i=0; i<n; ++i)	{
		scanf ("%d", &x);
		if (x != 0)	b[tot2++] = x;
	}
	int s = 0;
	for (; s<tot2; ++s)	{
		if (b[s] == a[0])	break;
	}
	bool flag = true;
	for (int i=0; i<tot1; ++i)	{
		if (b[(i+s)%(n-1)] != a[i])	{
			flag = false;	break;
		}
	}
	if (flag)	puts ("YES");
	else	puts ("NO");

	return 0;
}

数位DP C - XOR Equation

题意：求多少对(a, b)满足a + b == s && a ^ b == x

分析：ｓ和ｘ拆分成二进制，dp[i][4][0/1]第一维是二进制长度，第二维是ａ和ｂ二进制下第ｉ位的方案（０１，１０，００，１１），第三维是是否第ｉ位会进位，递推一下能够在ｌｏｇ（Ｎ）解决．

#include <bits/stdc++.h>

typedef long long ll;
int s[41], x[41];
ll dp[41][4][2];

void updata(int id)	{
	if (s[id] == x[id])	{
		if (s[id] == 0)	{	//0 0
			if (id == 0)	{
				dp[id][2][0] = dp[id][3][1] = 1;
			}
			else	{
				for (int j=0; j<4; ++j)	{
					dp[id][2][0] += dp[id-1][j][0];
					dp[id][3][1] += dp[id-1][j][0];
				}
			}
		}
		else	{			//1 1
			if (id == 0)	{
				dp[id][0][0] = dp[id][1][0] = 1;
			}
			else	{
				for (int j=0; j<4; ++j)	{
					dp[id][0][0] += dp[id-1][j][0];
					dp[id][1][0] += dp[id-1][j][0];
				}
			}
		}
	}
	else	{
		if (s[id] == 0)	{	//0 1
			for (int j=0; j<4; ++j)	{
				dp[id][0][1] += dp[id-1][j][1];
				dp[id][1][1] += dp[id-1][j][1];
			}
		}
		else	{			//1 0
			for (int j=0; j<4; ++j)	{
				dp[id][2][0] += dp[id-1][j][1];
				dp[id][3][1] += dp[id-1][j][1];
			}
		}
	}
}

int main()	{
	ll S, X;	std::cin >> S >> X;
	if (X == 0)	{
		puts ("1");	return 0;
	}
	bool same = false;
	if (S == X)	same = true;
	int n = 0, m = 0;
	while (S)	{
		if (S & 1)	s[n++] = 1;
		else	s[n++] = 0;
		S >>= 1;
	}
	while (X)	{
		if (X & 1)	x[m++] = 1;
		else	x[m++] = 0;
		X >>= 1;
	}
	int len = std::max (n, m);
	for (int i=0; i<len; ++i)	{
		updata (i);
	}
	ll ans = 0;
	for (int i=0; i<4; ++i)	{
		ans += dp[len-1][i][0];
	}
	if (same && ans >= 2)	ans -= 2;
	std::cout << ans << '\n';

	return 0;
}

树状数组　D - Factory Repairs

单点更新以及区间求和，不过不知道该点是ｋ天前还是后，树状数组开二维．

import java.io.*;
import java.util.*;

public class Main	{
	static int[][] C;
	static int[] A;
	static int n;
	static void updata(int i, int j, int x)	{
		while (i <= n)	{
			C[i][j] += x;
			i += i & -i;
		}
	}
	static int query(int i, int j)	{
		int ret = 0;
		while (i > 0)	{
			ret += C[i][j];
			i -= i & -i;
		}
		return ret;
	}
	static int min(int a, int b)	{
		if (a < b)	return a;
		else	return b;
	}
	public static void main(String[] args)	{
		Scanner cin = new Scanner (new BufferedInputStream (System.in));
		n = cin.nextInt ();
		int k = cin.nextInt ();
		int a = cin.nextInt ();
		int b = cin.nextInt ();
		int q = cin.nextInt ();
		C = new int[n+5][2];
		A = new int[n+5];
		for (int i=0; i<q; ++i)	{
			int op = cin.nextInt ();
			if (op == 1)	{
				int d = cin.nextInt ();
				int e = cin.nextInt ();
				int tmp = A[d];	A[d] += e;
				updata (d, 0, min (b, A[d]) - min (b, tmp));
				updata (d, 1, min (a, A[d]) - min (a, tmp));
			}
			else	{
				int p = cin.nextInt ();
				int ans = query (p - 1, 0) + query (n, 1) - query (p+k-1, 1);
				System.out.println (ans);
			}
		}
	}
}

贪心＋双端队列　E - Package Delivery

题意：ｍ个加油站，每单位油有价格，汽车有油的容量，初始满油，问到终点最少消费多少．

分析： 假设起点也算是一个加油站，免费，且汽车初始空油，那么每一个加油站最多能给汽车充ｎ单位油，那么到下一个加油站要选择之前加油站价格最少的加油，这里用双端队列维护最低价格的加油站．

#include <bits/stdc++.h>

typedef long long ll;
const int N = 2e5 + 5;
int dque[N];
std::pair<int, int> station[N];
int d, n, m;

int main()	{
	scanf ("%d%d%d", &d, &n, &m);
	for (int i=0; i<m; ++i)	{
		scanf ("%d%d", &station[i].first, &station[i].second);
	}
	station[m++] = std::make_pair (0, 0);
	station[m++] = std::make_pair (d, 0);
	std::sort (station, station+m);
	for (int i=0; i<m-1; ++i)	{
		if (station[i].first + n < station[i+1].first)	{
			puts ("-1");	return 0;
		}
	}
	int qs = 0, qe = 0, now = 0;	ll ans = 0;
	for (int i=0; i<m; ++i)	{
        int x = station[i].first, p = station[i].second;
        while (qs < qe && station[dque[qs]].first + n < x)	{
			ans += 1ll * (station[dque[qs]].first + n - now) * station[dque[qs]].second;
			now = station[dque[qs]].first + n;
			qs++;
        }
        ans += 1ll * (x - now) * station[dque[qs]].second;
        now = x;
        while (qs < qe && station[dque[qe-1]].second > p)	{
			qe--;
        }
        dque[qe++] = i;
	}
	std::cout << ans << '\n';

	return 0;
}

树形ＤＰ F - Preorder Test
题意：选择一条ｋ长的ＤＦＳ先序遍历路径，使得其中的最小值最大
分析：二分枚举最小值，然后枚举每个点出发的大于最小值的路径长度．求路径长度用到树形ＤＰ
首先ｄｏｗｎ［ｕ］表示ｕ的子树下最多能走的路径长度，然后第二次ｄｐ［ｕ］表示以ｕ为根节点下最多能走的路径长度，ｕｐ是ｕ以上完整的最大长度

#include <bits/stdc++.h>

const int N = 2e5 + 5;
struct DP {
	int sum, mx;
	DP operator + (const DP &rhs) const {
		return DP {sum + rhs.sum, std::max (mx, rhs.mx)};
	}
};
int a[N], total[N], down[N], dp[N];
bool good[N];
std::vector<int> edge[N];
int n, k;

void DFS(int u, int fa)	{
	total[u] = 1;
	int sum = 0, mx = 0;
	for (auto v: edge[u])	{
		if (v == fa)	continue;
		DFS (v, u);
		total[u] += total[v];
		if (down[v] == total[v]) {
			sum += total[v];
		} else {
			mx = std::max (mx, down[v]);
		}
	}
	down[u] = sum + mx + 1;
	if (!good[u])	down[u] = 0;
}

void DFS2(int u, int fa, int up)	{　　　　//learn from tourist
	std::vector<int> children;
	int sum = 0, mx = 0;
	for (auto v: edge[u])	{
		if (v == fa)	continue;
		if (down[v] == total[v]) {
			sum += down[v];
		} else {
			mx = std::max (mx, down[v]);
		}
		children.push_back (v);
	}
	if (up == n - total[u]) {
		sum += up;
	} else {
		mx = std::max (mx, up);
	}
	dp[u] = sum + mx + 1;
	if (!good[u])	dp[u] = 0;

	int sz = children.size ();
	if (sz == 0)	return ;
	std::vector<DP> predown (sz + 1);
	predown[0] = {0, 0};
	for (int i=0; i<sz; ++i)	{
		int v = children[i];
		predown[i+1] = predown[i];
		if (down[v] == total[v]) {
			predown[i+1].sum += down[v];
		} else {
			predown[i+1].mx = std::max (predown[i+1].mx, down[v]);
		}
	}
	std::vector<DP> sufdown (sz + 1);
	sufdown[sz] = {0, 0};
	for (int i=sz-1; i>=0; --i)	{
		int v = children[i];
		sufdown[i] = sufdown[i+1];
		if (down[v] == total[v]) {
			sufdown[i].sum += down[v];
		} else {
			sufdown[i].mx = std::max (sufdown[i].mx, down[v]);
		}
	}
	for (int i=0; i<sz; ++i) {
		int v = children[i];
		DP now = predown[i] + sufdown[i+1];
		if (up == n - total[u]) {
			now.sum += up;
		} else {
			now.mx = std::max (now.mx, up);
		}
		int new_up = now.sum + now.mx + 1;
		if (!good[u]) new_up = 0;
		DFS2 (v, u, new_up);
	}
}

bool check(int v)	{
	for (int i=1; i<=n; ++i)	good[i] = (a[i] >= v);
	DFS (1, 0);
	DFS2 (1, 0, 0);
	for (int i=1; i<=n; ++i)	{
		if (dp[i] >= k)	return true;
	}
	return false;
}

int main()	{
	scanf ("%d%d", &n, &k);
	for (int i=1; i<=n; ++i)	{
		scanf ("%d", a+i);
	}
	for (int u, v, i=1; i<n; ++i)	{
		scanf ("%d%d", &u, &v);
		edge[u].push_back (v);
		edge[v].push_back (u);
	}
	int low = 0, high = 1000005;
	while (low < high)	{
		int mid = (low + high + 1) >> 1;
		if (check (mid))	{
			low = mid;
		} else {
			high = mid - 1;
		}
	}
	printf ("%d\n", low);

	return 0;
}