AtCoder Beginner Contest 187

A Large Digits

int n;
int main()
{
	IOS;
	int a, b, resa = 0, resb = 0;
	cin >> a >> b;
	while(a) resa += a % 10, a /= 10;
	while(b) resb += b % 10, b /= 10;
	cout << max(resa, resb) << endl;
	return 0;
}

B Gentle Pairs

int n;
int x[N], y[N];
int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++ i) scanf("%d%d", &x[i], &y[i]);
	int res = 0;
	for(int i = 1; i <= n; ++ i)
		for(int j = i + 1; j <= n; ++ j)
		{
			if(x[i] - x[j] == 0) continue;
			if(x[i] - x[j] > 0)
			{
				if(y[i] - y[j] >= x[j] - x[i] && y[i] - y[j] <= x[i] - x[j])
				res ++;
			}
			if(x[i] - x[j] < 0)
			{
				if(y[i] - y[j] <= x[j] - x[i] && y[i] - y[j] >= x[i] - x[j])
					res ++;
			}
		}
	printf("%d\n", res);
}

C 1-SAT

int n;
map mp;
string str;
int main()
{
	IOS;
	cin >> n;
	for(int i = 1; i <= n; ++ i)
	{
		cin >> str;
		if(str[0] == '!' && mp.count(str.substr(1)))
		{
			cout << str.substr(1) << endl;
			return 0;
		}
		if(str[0] != '!' && mp.count("!" + str))
		{
			cout << str << endl;
			return 0;
		}
		mp[str] = 1;
	}
	puts("satisfiable");
	return 0;
}

D Choose Me

不选时\(A += a_i\), \(B += 0\), 选时 \(A += 0\), \(B += a_i + b_i\)

先考虑都不选，每选择一个相当于从\(A\)中减去一个\(a_i\)，\(B\)加一个\(a_i + b_i\)，因为是比较大小关系，所以等价于\(A\)不动，\(B + 2 \times a_i + b_i\)

int n;
struct zt
{
	int a, b;
	LL c;
}t[N];
bool cmp(zt a, zt b)
{
	return a.c > b.c;
}
int main()
{
	scanf("%d", &n);
	LL sum = 0, res = 0;
	for(int i = 1; i <= n; ++ i)
	{
		scanf("%d%d", &t[i].a, &t[i].b);
		t[i].c = t[i].a * 2 + t[i].b;
		sum += t[i].a;
	}
	sort(t + 1, t + n + 1, cmp);
	for(int i = 1; i <= n; ++ i)
	{
		res += t[i].c;
		if(res > sum) { printf("%d\n", i); break;}
	}
	return 0;
}

E Through Path

树上差分

以1号点为根，预处理每个点的深度.

若对于深度大的点所在分支（即它的子树）全部\(+x\),只需要对该点\(+x\).

若对于深度小的点所在分支全部\(+x\)，可以看成所有点\(+x\),深度大的点所在子树\(-x\)，故维护一个所有点的增加值，并对深度大的点\(-x\).

最后遍历一遍子树把标记下传即可.

int n, m;
struct Edge
{
	int to, nxt;
}line[N * 2];
int fist[N], idx;
int dep[N];
LL d[N], res;
void add(int x, int y)
{
	line[++ idx] = (Edge){y, fist[x]};
	fist[x] = idx;
}
void dfs(int u, int far)
{
	for(int i = fist[u]; i != -1; i = line[i].nxt)
	{
		int v = line[i].to;
		if(v == far) continue;
		dep[v] = dep[u] + 1;
		dfs(v, u);
	}
}
void dfs2(int u, int far)
{
	for(int i = fist[u]; i != -1; i = line[i].nxt)
	{
		int v = line[i].to;
		if(v == far) continue;
		d[v] += d[u];
		dfs2(v, u);
	}
}
int main()
{
	memset(fist, -1, sizeof fist);
	scanf("%d", &n);
	for(int i = 1; i < n; ++ i)
	{
		int a, b;
		scanf("%d%d", &a, &b);
		add(a, b);
		add(b, a);
	}
	dfs(1, -1);
	scanf("%d", &m);
	for(int i = 1; i <= m; ++ i)
	{
		int t, e, x;
		scanf("%d%d%d", &t, &e, &x);
		int a = line[2 * e].to, b = line[2 * e - 1].to;
		if(dep[a] < dep[b] && t == 2) d[b] += x;
		if(dep[a] > dep[b] && t == 1) d[a] += x;
		if(dep[a] < dep[b] && t == 1) { d[b] -= x; res += x; }
		if(dep[a] > dep[b] && t == 2) { d[a] -= x; res += x; }
	}
	dfs2(1, -1);
	for(int i = 1; i <= n; ++ i) printf("%lld\n", d[i] + res);
	return 0;
}

F Close Group

令f[i]表示状态i划分的团的个数集合中的最小值.

枚举i的子集s， f[i] = min(f[i], f[i - s] + f[s])

int n, m;
int conn[1 << N];
int f[1 << N];
int main()
{
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= m; ++ i)
	{
		int a, b;
		scanf("%d%d", &a, &b);
		a -- , b -- ;
		conn[a] |= (1 << b);
		conn[b] |= (1 << a);
	}
	for(int i = 0; i < n; ++ i) conn[i] |= (1 << i);
	memset(f, 0x3f, sizeof f);
	f[0] = 0;
	for(int i = 1; i < (1 << n); ++ i)
	{
		int flag = 1;
		for(int j = 0; j < n; ++ j)
			if(i & (1 << j) && (conn[j] & i) != i)
			{
				flag = 0;
				break;
			}
		if(flag) f[i] = 1;
		for(int s = i; s; s = (s - 1) & i)
			f[i] = min(f[i], f[i - s] + f[s]);
	}
	printf("%d\n", f[(1 << n) - 1]);
	return 0;
}

2021.1.21