Atcoder Grand Contest 004 题解

A - Divide a Cuboid

如果一边是偶数，肯定一刀切一半最优，否则看一下切出来的差就是另外两边的乘积。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
int A, B, C; 
 
int main()
{
	giii(A, B, C);
	if (A % 2 == 0 || B % 2 == 0 || C % 2 == 0) puts("0");
	else printf("%lld
", min(1LL * A * B, min(1LL * B * C, 1LL * A * C)));
	return 0;
}

　　

B - Colorful Slimes

枚举转了y次，那么一个点被造出来的花费就是min(a[i-y]...a[i])，最后加上x*y对所有情况取min就好了。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
int n, a[2010], x, mn[2010][2010];
 
long long ans;
 
int main()
{
	gii(n, x);
	for (int i = 1; i <= n; ++i) gi(a[i]), ans += a[i];
	for (int i = 1; i <= n; ++i)
	{
		mn[i][i] = a[i];
		for (int j = i + 1; j <= n; ++j)
			mn[i][j] = min(mn[i][j - 1], a[j]);
	}
	for (int i = 0; i <= n; ++i)
	{
		long long ret = 1LL * x * i;
		for (int j = 1; j <= n; ++j)
		{
			int k = j - i;
			if (k < 1) k += n, ret += min(mn[k][n], mn[1][j]);
			else ret += mn[k][j];
		}
		ans = min(ans, ret);
	}
	printf("%lld
", ans);
	return 0;
}

　　

C - AND Grid

两边摆成类似正反E字交错放，重复地方两个都放就好了。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
int n, m;
 
char str[510][510];
 
int main()
{
	gii(n, m);
	for (int i = 1; i <= n; ++i) scanf("%s", str[i] + 1);
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 1; j <= m; ++j)
		{
			if (((j == 1 || (i & 1)) && j != m) || str[i][j] == '#')
				putchar('#');
			else
				putchar('.');
		}
		putchar('
');
	}
	for (int i = 1; i <= n; ++i)
	{
		for (int j = 1; j <= m; ++j)
		{
			if (((j == m || !(i & 1)) && j != 1) || str[i][j] == '#')
				putchar('#');
			else
				putchar('.');
		}
		putchar('
');
	}
}

　　

D - Teleporter

首先，1号必须连自己，因为如果1有在一个环上，环上每个点走k次结果都不一样，其它也只能改为1号最优。

那么，省下就是一棵树了，bfs一遍即可。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
const int N = 1e5 + 10;
 
int n, k, a[N]; 
 
int ans = 0;
 
int deg[N], dep[N];
 
int main()
{
	gii(n, k); --k;
	for (int i = 1; i <= n; ++i) gi(a[i]);
	if (a[1] != 1) ++ans, a[1] = 1;
	for (int i = 1; i <= n; ++i)
		++deg[a[i]];
	static int q[N]; int l = 0, r = 0;
	for (int i = 1; i <= n; ++i)
	{
		if (!deg[i]) q[r++] = i;
	}
	while (l < r)
	{
		int u = q[l++];
		--deg[a[u]];
		if (!deg[a[u]]) q[r++] = a[u];
		if (dep[u] == k && a[u] != 1) a[u] = 1, ++ans;
		else dep[a[u]] = max(dep[a[u]], dep[u] + 1);
	}
	printf("%d
", ans);
	return 0;
}

　　

E - Salvage Robots

能走的区域是一个矩形，但是能取到的就长得奇形怪状了，所以我们就写一个四位dp，算算边界就好了。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
int H, W;
 
char str[110][110];
 
int cnt[110][110];
 
int sum(int x1, int y1, int x2, int y2)
{
	x1 = max(x1, 1), x1 = min(x1, H);
	x2 = max(x2, 1), x2 = min(x2, H);
	y1 = max(y1, 1), y1 = min(y1, W);
	y2 = max(y2, 1), y2 = min(y2, W);
	if (x1 > x2) return 0;
	if (y1 > y2) return 0;
	//cerr << x1 << ", " << y1 << " and " << x2 << ", " << y2 << endl;
	return cnt[x2][y2] + cnt[x1 - 1][y1 - 1] - cnt[x1 - 1][y2] - cnt[x2][y1 - 1];
}
 
int f[2][110][110][110];
 
int main()
{
	gii(H, W);
	for (int i = 1; i <= H; ++i)
		scanf("%s", str[i] + 1);
	int x = 0, y = 0;
	for (int i = 1; i <= H; ++i)
		for (int j = 1; j <= W; ++j)
			if (cnt[i][j] = (str[i][j] == 'o'), str[i][j] == 'E')
				x = i, y = j;
	for (int i = 1; i <= H; ++i)
		for (int j = 1; j <= W; ++j)
			cnt[i][j] += cnt[i - 1][j] + cnt[i][j - 1] - cnt[i - 1][j - 1];
	int ans = 0;
	for (int l = 0; x + l <= H; ++l)
		for (int r = 0; x - r > 0; ++r)
			for (int u = 0; y + u <= W; ++u)
				for (int d = 0; y - d > 0; ++d)
				{
					int L = l & 1;
					f[L][r][u][d] = 0;
					int left = x - r, right = x + l;
					int up = y - d, down = y + u;
					left = max(1 + l, left);
					right = min(H - r, right);
					up = max(up, 1 + u);
					down = min(down, W - d);
					if (l && l + r + x <= H) f[L][r][u][d] = max(f[L][r][u][d], f[L ^ 1][r][u][d] + sum(x + l, up, x + l, down));
					if (r && l + r <= x - 1) f[L][r][u][d] = max(f[L][r][u][d], f[L][r - 1][u][d] + sum(x - r, up, x - r, down));
					if (u && u + d + y <= W) f[L][r][u][d] = max(f[L][r][u][d], f[L][r][u - 1][d] + sum(left, y + u, right, y + u));
					if (d && u + d <= y - 1) f[L][r][u][d] = max(f[L][r][u][d], f[L][r][u][d - 1] + sum(left, y - d, right, y - d));
					//cerr << l << ", " << r << ", " << u << ", " << d << " : " << f[l][r][u][d] << endl; 
					ans = max(ans, f[L][r][u][d]);
				}
	printf("%d
", ans);
	return 0;
}

　　

F - Namori

我们首先可以发现一个性质，黑色只能成对出现而且，这两个点距离为奇数，次数是距离。

那么树就很好写了，只要一个启发式合并，求深度的奇偶性。

环套树我们把它缩成环，然后每个附上权值d[i]，偶环就是一个负载平衡问题，奇环我们可以断开一条边，因为那条边的作用是改变奇偶用的，所以我们扫一遍就能知道会改变多少次。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
	char ch;
	int x, a;
	while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
	if (ch == '-') ch = getchar(), a = -1;
	else a = 1;
	x = ch - '0';
	while (ch = getchar(), ch >= '0' && ch <= '9')
		x = (x << 1) + (x << 3) + ch - '0';
	return a * x;
}
 
const int N = 1e5 + 10;
 
VI edge[N];
 
int n, m;
 
namespace task1
{
	set<int> e[N][2];
	
	int fa[N], dep[N], check_cnt;
	
	void dfs1(int u)
	{
		dep[u] = dep[fa[u]] + 1;
		e[u][dep[u] & 1].insert(u);
		if (dep[u] & 1) ++check_cnt;
		else --check_cnt;
		for (auto v : edge[u])
		{
			if (v == fa[u]) continue;
			fa[v] = u;
			dfs1(v);
		}
	}
	
	long long ans;
	
	void dfs2(int u)
	{
		for (auto v : edge[u])
		{
			if (v == fa[u]) continue;
			dfs2(v);
			if (e[u][1].size() < e[v][1].size())
				e[u][1].swap(e[v][1]);
			for (auto x : e[v][1])
				e[u][1].insert(x);
			if (e[u][0].size() < e[v][0].size())
				e[u][0].swap(e[v][0]);
			for (auto x : e[v][0])
				e[u][0].insert(x);
		}
		if (e[u][1].size() && e[u][0].size())
		{
			int s = min(e[u][1].size(), e[u][0].size());
			set<int>::iterator it1 = e[u][1].begin(), it2 = e[u][0].begin();
			static PII stk[N]; int tp = 0;
			for (int i = 1; i <= s; ++i)
			{
				int v1 = *it1;
				int v2 = *it2;
				stk[++tp] = mp(v1, v2);
				ans += dep[v1] + dep[v2] - (dep[u] << 1);
				//cerr << v1 << ", " << v2 << ", " << u << endl;
				++it1, ++it2;
			}
			for (int i = 1; i <= tp; ++i)
				e[u][1].erase(stk[i].fi), e[u][0].erase(stk[i].se);
		}
	}
	
	void solve()
	{
		dfs1(1);
		if (check_cnt)
		{
			puts("-1");
			return;
		}
		dfs2(1);
		printf("%lld
", ans);
	}
}
 
namespace task2
{
	bool vis[N], cir[N];
	int use_fa[N];
	vector<int> circle;
	bool exit_flag;
	void dfs1(int u)
	{
		vis[u] = 1;
		for (auto v : edge[u])
		{
			if (exit_flag) return;
			if (use_fa[u] == v) continue;
			if (vis[v])
			{
				while (u != v) circle.pb(u), cir[u] = 1, u = use_fa[u];
				circle.pb(v), cir[v] = 1;
				exit_flag = 1;
				return;
			}
			use_fa[v] = u;
			dfs1(v);
		}
	}
	
	int dep[N], fa[N];
	
	set<int> e[N][2];
	
	void dfs2(int u)
	{
		dep[u] = dep[fa[u]] + 1;
		e[u][dep[u] & 1].insert(u);
		for (auto v : edge[u])
		{
			if (v == fa[u]) continue;
			if (cir[v]) continue;
			fa[v] = u;
			dfs2(v);
		}
	}
	
	long long ans;
	
	void dfs3(int u)
	{
		for (auto v : edge[u])
		{
			if (v == fa[u]) continue;
			if (cir[v]) continue;
			dfs3(v);
			if (e[u][1].size() < e[v][1].size())
				e[u][1].swap(e[v][1]);
			for (auto x : e[v][1])
				e[u][1].insert(x);
			if (e[u][0].size() < e[v][0].size())
				e[u][0].swap(e[v][0]);
			for (auto x : e[v][0])
				e[u][0].insert(x);
		}
		if (e[u][1].size() && e[u][0].size())
		{
			int s = min(e[u][1].size(), e[u][0].size());
			set<int>::iterator it1 = e[u][1].begin(), it2 = e[u][0].begin();
			static PII stk[N]; int tp = 0;
			for (int i = 1; i <= s; ++i)
			{
				int v1 = *it1;
				int v2 = *it2;
				stk[++tp] = mp(v1, v2);
				ans += dep[v1] + dep[v2] - (dep[u] << 1);
				//cerr << v1 << ", " << v2 << ", " << u << endl;
				++it1, ++it2;
			}
			for (int i = 1; i <= tp; ++i)
				e[u][1].erase(stk[i].fi), e[u][0].erase(stk[i].se);
		}
	}
	
	int d[N];
	
	void solve()
	{
		dfs1(1);
		for (auto root : circle)
		{
			//cerr << root << endl;
			dfs2(root), dfs3(root);
			if (e[root][1].size())
			{
				for (auto v : e[root][1])
					++d[root], ans += dep[v] - 1;
			}
			else if (e[root][0].size())
			{
				for (auto v : e[root][0])
					--d[root], ans += dep[v] - 1;
			}
		}
		int vtx = SZ(circle);
		//cerr << "ok!" << endl;
		if (vtx % 2 == 0)
		{
			int sum = 0;
			for (int i = 0; i < vtx; ++i)
			{
				if (i & 1) d[circle[i]] = -d[circle[i]];
				sum += d[circle[i]];
			}
			if (sum) 
			{
				puts("-1");
				return;
			}
			static vector<int> t;
			t.pb(-d[circle[0]]);
			for (int i = 1; i < vtx; ++i)
			{
				d[circle[i]] += d[circle[i - 1]];
				t.pb(-d[circle[i]]);
			}
			sort(t.begin(), t.end());
			int v = t[SZ(t) >> 1];
			for (auto x : t)
				ans += abs(x - v);
			printf("%lld
", ans);
		}
		else
		{
			int sum = 0;
			for (int i = 0; i < vtx; ++i)
			{
				if (i & 1) d[circle[i]] = -d[circle[i]];
				sum += d[circle[i]];
			}
			if (sum % 2 != 0)
			{
				puts("-1");
				return;
			}
			ans += abs(sum / 2);
			d[circle[0]] -= sum / 2;
			d[circle[vtx - 1]] += sum / 2;
			for (int i = 0; i < vtx - 1; ++i)
			{
				ans += abs(d[circle[i]]);
				d[circle[i + 1]] += d[circle[i]];
			}
			printf("%lld
", ans);
		} 
	}
}
 
int main()
{
	gii(n, m);
	for (int i = 1; i <= m; ++i)
	{
		int u, v;
		gii(u, v);
		edge[u].push_back(v);
		edge[v].push_back(u);
	}
	if (m == n - 1)
	{
		task1::solve();
		return 0;
	}
	task2::solve();
	return 0;
}

　　

1. //waz
2. #include<bits/stdc++.h>
3.  
4. usingnamespace std;
5.  
6. #define mp make_pair
7. #define pb push_back
8. #definefi first
9. #define se second
10. #define ALL(x)(x).begin(),(x).end()
11. #define SZ(x)((int)((x).size()))
12.  
13. typedef pair<int,int> PII;
14. typedef vector<int> VI;
15. typedeflonglong int64;
16. typedefunsignedintuint;
17. typedefunsignedlonglong uint64;
18.  
19. #define gi(x)((x)= F())
20. #define gii(x, y)(gi(x), gi(y))
21. #define giii(x, y, z)(gii(x, y), gi(z))
22.  
23. int F()
24. {
25. char ch;
26. int x, a;
27. while(ch = getchar(),(ch <'0'|| ch >'9')&& ch !='-');
28. if(ch =='-') ch = getchar(), a =-1;
29. else a =1;
30. x = ch -'0';
31. while(ch = getchar(), ch >='0'&& ch <='9')
32. x =(x <<1)+(x <<3)+ ch -'0';
33. return a * x;
34. }
35.  
36. constint N =1e5+10;
37.  
38. int n, k, a[N];
39.  
40. int ans =0;
41.  
42. int deg[N], dep[N];
43.  
44. int main()
45. {
46. gii(n, k);--k;
47. for(int i =1; i <= n;++i) gi(a[i]);
48. if(a[1]!=1)++ans, a[1]=1;
49. for(int i =1; i <= n;++i)
50. ++deg[a[i]];
51. staticint q[N];int l =0, r =0;
52. for(int i =1; i <= n;++i)
53. {
54. if(!deg[i]) q[r++]= i;
55. }
56. while(l < r)
57. {
58. int u = q[l++];
59. --deg[a[u]];
60. if(!deg[a[u]]) q[r++]= a[u];
61. if(dep[u]== k && a[u]!=1) a[u]=1,++ans;
62. else dep[a[u]]= max(dep[a[u]], dep[u]+1);
63. }
64. printf("%d ", ans);
65. return0;
66. }