ICPC 2019-2020 North-Western Russia Regional Contest 部分题解

E. Equidistant

题意：有一颗树，有$m$个特殊点，问是否存在一个点到$m$个特殊点的距离相等

解法：换根$dp$秒之，设$mx[u],mn[u]$分别为$u$子树中距离$u$最远的特殊点的距离和最近的特殊点的距离，首先一次$dfs$搞定树$dp$，接下来就是直接换根了，当一个点$u$为根时$mx[u]=mn[u]$，那么这个点一定是合法的点

#include
#define pi pair
#define mk make_pair
using namespace std;
const int maxn = 2e5 + 10, inf = 1e9;
vector<int> G[maxn];
vector<pi> s[maxn];
int mx[maxn], mn[maxn], cat[maxn], jie, n;
void dfs(int u, int fa) {
	if (cat[u])
		mx[u] = mn[u] = 0;
	else 
		mx[u] = -inf, mn[u] = inf;
	for (auto v : G[u])
		if (v != fa) {
			dfs(v, u);
			mx[u] = max(mx[u], mx[v] + 1);
			mn[u] = min(mn[u], mn[v] + 1);
		}
}
void dfs2(int u, int fa) {
	if (jie)
		return;
	if (cat[u])
		mx[u] = mn[u] = 0;
	else 
		mx[u] = -inf, mn[u] = inf;
	s[u].push_back(mk(mx[u],mn[u]));
	for (auto v : G[u]) {
		mx[u] = max(mx[u], mx[v] + 1);
		mn[u] = min(mn[u], mn[v] + 1);
		s[u].push_back(mk(mx[u], mn[u]));
	}
	if (mx[u] == mn[u]) {
		jie = u;
		return;
	}
	int Mx = -inf, Mn = inf;
	for (int i = G[u].size() - 1; ~i; i--) {
		int v = G[u][i];
		mx[u] = max(s[u][i].first, Mx);
		mn[u] = min(s[u][i].second, Mn);
		Mx = max(Mx, mx[v] + 1);
		Mn = min(Mn, mn[v] + 1);
		if (v != fa)
			dfs2(v, u);
	}
}
int main()
{
	int m, u, v;
	scanf("%d%d", &n, &m);
	for (int i = 1; i < n; i++) {
		scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for (int i = 1; i <= m; i++) {
		scanf("%d", &u);
		cat[u] = v;
	}
	dfs(1, 0);
	dfs2(1, 0);
	if (jie)
		printf("YES\n%d\n", jie);
	else 
		puts("NO");
}

H. High Load Database

题意：有$n$个东西重量总和不超过$1e6$，有$q$次询问容量为$x$的背包按顺序装东西至少需要多少个背包能装完所有东西

解法：我们预处理一下所有$size$的背包一次性能装几个东西，每次询问我们暴力去模拟一下即可，复杂度为调和级数

#include
using namespace std;
const int maxn = 1e6 + 5, N = 1e6;
int a[maxn], p[maxn], d[maxn], sum[maxn], mx, n;
int gao(int x) {
	if (x < mx)
		return -1;
	if (d[x])
		return d[x];
	int res = 0, i = 0;
	while (i < n) {
		res += x;
		res = min(res, N);
		i = p[res];
		res = sum[i];
		d[x]++;
	}
	return d[x];
}
int main()
{
	int q, x;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		mx = max(mx, a[i]);
		sum[i] = sum[i - 1] + a[i];
		for (int j = sum[i - 1]; j < sum[i]; j++)
			p[j] = i - 1;
	}
	for (int i = sum[n]; i <= N; i++)
		p[i] = n;
	scanf("%d", &q);
	while (q--) {
		scanf("%d", &x);
		if (gao(x) != -1)	
			printf("%d\n", gao(x));
		else 
			printf("Impossible\n");
	}
}

I. Ideal Pyramid

题意：给出$n$个东西的二维坐标和高度，你要建立一个高度最小的四棱锥，包含住这$n$个东西，要求四棱锥底面是个和$x$轴平行的正方形，四棱锥的四个面和二维平面呈45度角，求四棱锥中间坐标和最小的高度（整型数，如果为小数，向上取整）

解法：太南了，我们写个简单版的题：在$x$轴上有$n$个东西有高度，你要在$x$轴上建立一个高度最小的等腰直角三角形，斜边和$x$轴重合，要求三角形能覆盖住这$n$个东西。这个题你会写吧？那对于原题意，我们可以忽视$y$坐标转化成$x$轴上找小的等腰三角形包含住它们，忽视$x$坐标转化成$y$轴上找最小的等腰直角三角形包含住它们，结合起来就是答案

#include 
using namespace std;
const int maxn = 1005;
struct node
{
    int x, y, h;
} a[maxn];
int main()
{
    int n, L1 = 2e9, L2 = 2e9, R1 = -2e9, R2 = -2e9, x, y;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> a[i].x >> a[i].y >> a[i].h;
    for (int i = 1; i <= n; i++)
    {
        L1 = min(L1, a[i].x - a[i].h);
        R1 = max(R1, a[i].x + a[i].h);
        L2 = min(L2, a[i].y - a[i].h);
        R2 = max(R2, a[i].y + a[i].h);
    }
    x = (L1 + R1) / 2;
    y = (L2 + R2) / 2;
    int h = max(x - L1, y - L2);
    int h2 = max(R1 - x, R2 - y);
    h = max(h, h2);
    printf("%d %d %d\n", (L1 + R1) / 2, (L2 + R2) / 2, h);
}

J. Just the Last Digit

题意：有一个有向图，满足如果$u>=v$，那么$u$肯定不会有一条边走向$v$，现在告诉你任意点$u$到任意点$v$的路径数模$10$等于$a[u][v]$，让你恢复这个有向图$(n<=500)$

解法：假设$u,v$有一条单向边直接相连，那么$d[u][v]=1，$我们从大到小枚举$u$，然后从$u+1$到$n$枚举$v$，然后枚举$u,v$之间的点$k$，设$sum$为$u$到$v$的路径数，如果$d[u][k]=1$，那么$sum+=a[k][v]$，如果最后$sum!=a[u][v]$，那么$u,v$肯定有单向边

#include
using namespace std;
const int maxn = 505;
char s[maxn][maxn];
int a[maxn][maxn];
int main()
{
	int n;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) {
		scanf("%s", s[i] + 1);
		for (int j = 1; j <= n; j++)
			s[i][j] -= '0';
	}
	for (int i = n - 1; i; i--)
		for (int j = i + 1; j <= n; j++) {
			int cat = 0;
			for (int k = i + 1; k < j; k++)
				cat = (cat + s[k][j] * a[i][k]) % 10;
			if (cat != s[i][j])
				a[i][j] = 1;
		}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
			printf("%d", a[i][j]);
		puts("");
	}
}