JZOJ 2474. 【GDKOI 2021普及组DAY2】我的世界

题解

这题很明显发现一个点到另一个点，必然最多只有一个进入下界的点和一个出来的点

分类讨论入点和出点的位置

要么都在 \(u->lca\) 或都在 \(lca->v\) 或分别有一个

那就有一个倍增做法

维护最优入点和最优出点即可

但考场并没想到这种方法

反而只想到了没脑的倍增维护矩阵转移的方法

我们设 \(f_{u,0/1}\) 表示从它的儿子 \(v\) 过来的最优答案

把转移式子写成矩阵的形式，倍增维护即可

但我的常数太大了，沦为和暴力老哥同分

\(Code\)

#pragma GCC optimize(3)
#pragma GCC optimize("inline")
#pragma GCC optimize("Ofast")
#pragma GCC target("sse3","sse2","sse")
#pragma GCC diagnostic error "-std=c++14"
#pragma GCC diagnostic error "-fwhole-program"
#pragma GCC diagnostic error "-fcse-skip-blocks"
#pragma GCC diagnostic error "-funsafe-loop-optimizations"
#pragma GCC optimize("fast-math","unroll-loops","no-stack-protector","inline")
#include<cstdio>
#include<iostream>
#define LL long long
using namespace std;

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

inline void read(int &x)
{
	x = 0; int f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') f = (ch == '-' ? -1 : f), ch = getchar();
	while (ch >= '0' && ch <= '9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();
	x *= f;
}
inline void read(LL &x)
{
	x = 0; int f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') f = (ch == '-' ? -1 : f), ch = getchar();
	while (ch >= '0' && ch <= '9') x = (x<<3) + (x<<1) + ch - '0', ch = getchar();
	x *= f;
}

int h[N], tot;
struct edge{int to, nxt; LL w;}e[N << 1];
inline void add(int x, int y, LL z){e[++tot] = edge{y, h[x], z}, h[x] = tot;}

const LL INF = 1e18;
struct Matrix{
	LL a[3][3];
	inline Matrix()
	{
		for(register int i = 0; i < 3; i++)
			for(register int j = 0; j < 3; j++) a[i][j] = INF;
	}
	inline Matrix operator * (const Matrix &b)
	{
		Matrix c;
		for(register int i = 0; i < 3; i++)
			for(register int j = 0; j < 3; j++)
				for(register int k = 0; k < 3; k++)
				c.a[i][j] = min(c.a[i][j], a[i][k] + b.a[k][j]);
		return c;
	}
}f[N][20];
inline Matrix I()
{
	Matrix c;
	c.a[0][0] = c.a[1][1] = c.a[2][2] = 0;
	return c;
}
inline Matrix BOT(int u)
{
	Matrix c;
	c.a[0][0] = c.a[1][0] = 0, c.a[2][0] = a[u];
	return c;
}
inline Matrix Node(int u, int v, LL w)
{
	Matrix c;
	c.a[0][0] = 8 * w, c.a[0][1] = w + a[u] + a[v], c.a[0][2] = a[u] + w;
	c.a[1][0] = 8 * w, c.a[1][1] = w + a[u] + a[v], c.a[1][2] = a[u] + w;
	c.a[2][0] = 8 * w + a[u], c.a[2][1] = w + a[v], c.a[2][2] = w;
	return c;
}

int dep[N], anc[N][20], d[N];
void bfs()
{
	int head = 0, tail = 1;
	d[1] = 1;
	while (head < tail)
	{
		int x = d[++head];
		for(register int i = 1; i <= 18; i++)
		if (anc[x][i - 1]) anc[x][i] = anc[anc[x][i - 1]][i - 1], f[x][i] =  f[anc[x][i - 1]][i - 1] * f[x][i - 1];
		else break;
		for(register int i = h[x]; i; i = e[i].nxt)
		{
			int v = e[i].to;
			if (v == anc[x][0]) continue;
			anc[v][0] = x, f[v][0] = Node(x, v, e[i].w);
			dep[v] = dep[x] + 1, d[++tail] = v;
		}
	}
}
inline LL getans(int x, int y)
{
	if (dep[x] < dep[y]) swap(x, y);
	int deep = dep[x] - dep[y];
	Matrix retx = I(), rety = I(), Bx = BOT(x), By = BOT(y);
	for(register int i = 18; i >= 0; i--)
	if ((deep >> i) & 1) retx = f[x][i] * retx , x = anc[x][i];
	if (x == y)
	{
		retx = retx * Bx;
		return min(retx.a[0][0], retx.a[2][0] + a[x]);
	}
	for(register int i = 18; i >= 0; i--)
	if (anc[x][i] ^ anc[y][i])
	{
		retx = f[x][i] * retx , rety = f[y][i] * rety;
		x = anc[x][i], y = anc[y][i];
	}
	retx = f[x][0] * retx * Bx, rety = f[y][0] * rety * By;
	return min(min(retx.a[0][0] + rety.a[0][0], retx.a[2][0] + rety.a[2][0]),
		min(retx.a[0][0] + rety.a[2][0] + a[anc[x][0]], retx.a[2][0] + rety.a[0][0] + a[anc[x][0]]));
}

int main()
{
	freopen("minecraft.in", "r", stdin);
	freopen("minecraft.out", "w", stdout);
	read(n);
	for(register int i = 1; i <= n; i++) read(a[i]);
	int x, y; LL z;
	for(register int i = 1; i < n; i++) read(x), read(y), read(z), add(x, y, z), add(y, x, z);
	bfs();
	int q; read(q);
	for(; q; --q) read(x), read(y), printf("%lld\n", getans(x, y));
}