【BZOJ 4515】【SDOI 2016 Round1 Day1 T3】游戏

考场上写了lct，可惜当时对标记永久化的理解并不是十分深刻，导致调一个错误的程序调了4h+，最后这道题爆0了QwQ

现在写了树链剖分，用标记永久化的线段树维护轻重链，对于$s\rightarrow lca$，$lca\rightarrow t$分开讨论，把$a×dist+b$这个式子打开，提出常数项，发现是一个一次函数(也不是严格的一次函数，只不过有单调性就可以做了)。标记永久化线段树做一下，每个节点维护一个当前区间的最小值就可以统计答案了233

这次又手残把“+”打成“*”调了一上午TwT，肉眼对拍大法好！

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int N = 100005;
const LL inf = 123456789123456789LL;
void read(int &k) {
	k = 0; int fh = 1; char c = getchar();
	for(; c < '0' || c > '9'; c = getchar())
		if (c == '-') fh = -1;
	for(; c >= '0' && c <= '9'; c = getchar())
		k = (k << 1) + (k << 3) + c - '0';
	k = k * fh;
}

struct node {
	int nxt, to, w;
} E[N << 1];
int n, m, cnt = 0, point[N], fa[N], top[N], son[N], sz[N], pos[N], wt[N], K[N << 2];
LL B[N << 2], val[N << 2], di[N], ans;
bool p[N << 2];
void ins(int x, int y, int z) {E[++cnt].nxt = point[x]; E[cnt].to = y; E[cnt].w = z; point[x] = cnt;}
void _(int x) {
	sz[x] = 1;
	for(int tmp = point[x]; tmp; tmp = E[tmp].nxt)
		if (E[tmp].to != fa[x]) {
			fa[E[tmp].to] = x;
			di[E[tmp].to] = di[x] + E[tmp].w;
			_(E[tmp].to);
			sz[x] += sz[E[tmp].to];
			if (sz[E[tmp].to] > sz[son[x]]) son[x] = E[tmp].to;
		}
}
void __(int x) {
	pos[x] = ++cnt; wt[cnt] = x;
	if (son[x]) {top[son[x]] = top[x]; __(son[x]);}
	for(int tmp = point[x]; tmp; tmp = E[tmp].nxt)
		if (E[tmp].to != fa[x] && E[tmp].to != son[x])
			{top[E[tmp].to] = E[tmp].to; __(E[tmp].to);}
}
int LCA(int x, int y) {
	for(; top[x] != top[y]; x = fa[top[x]])
		if (di[top[x]] < di[top[y]]) swap(x, y);
	return di[x] < di[y] ? x : y;
}
void pushup(int rt, int l, int r) {
	if (l < r) val[rt] = min(val[rt << 1], val[rt << 1 | 1]); else val[rt] = inf;
	if (p[rt]) val[rt] = min(val[rt], min(di[wt[l]] * K[rt], di[wt[r]] * K[rt]) + B[rt]);
}
void Build(int rt, int l, int r) {
	val[rt] = inf;
	if (l == r) return;
	int mid = (l + r) >> 1;
	Build(rt << 1, l, mid);
	Build(rt << 1 | 1, mid + 1, r);
}

void update(int rt, int l, int r, int a, LL b) {
	if (!p[rt]) {
		p[rt] = 1; K[rt] = a; B[rt] = b;
	} else {
		LL newl = a * di[wt[l]] + b, newr = a * di[wt[r]] + b, befl = K[rt] * di[wt[l]] + B[rt], befr = K[rt] * di[wt[r]] + B[rt];
		int mid = (l + r) >> 1;
		if (newl <= befl && newr <= befr) {
			K[rt] = a; B[rt] = b;
		} else
			if (newl >= befl && newr >= befr)
				return;
			else
				if (a < K[rt]) {
					double tmp = (b - B[rt]) / (K[rt] - a);
					if (tmp <= di[wt[mid]]) {
						update(rt << 1, l, mid, K[rt], B[rt]);
						K[rt] = a; B[rt] = b;
					} else
						update(rt << 1 | 1, mid + 1, r, a, b);
				} else {
					double tmp = (B[rt] - b) / (a - K[rt]);
					if (tmp > di[wt[mid]]) {
						update(rt << 1 | 1, mid + 1, r, K[rt], B[rt]);
						K[rt] = a; B[rt] = b;
					} else
						update(rt << 1, l, mid, a, b);
				}
	}
	pushup(rt, l, r);
}

void inc(int rt, int l, int r, int L, int R, int a, LL b) {
	if (L <= l && r <= R) {update(rt, l, r, a, b); return;}
	int mid = (l + r) >> 1;
	if (L <= mid) inc(rt << 1, l, mid, L, R, a, b);
	if (R > mid) inc(rt << 1 | 1, mid + 1, r, L, R, a, b);
	pushup(rt, l, r);
}
void Q(int rt, int l, int r, int L, int R) {
	if (L == l && r == R) {ans = min(ans, val[rt]); return;}
	if (p[rt]) ans = min(ans, min(K[rt] * di[wt[L]], K[rt] * di[wt[R]]) + B[rt]);
	int mid = (l + r) >> 1;
	if (R <= mid) Q(rt << 1, l, mid, L, R);
	else if (L > mid) Q(rt << 1 | 1, mid + 1, r, L, R);
	else {Q(rt << 1, l, mid, L, mid); Q(rt << 1 | 1, mid + 1, r, mid + 1, R);}
}

int main() {
	read(n); read(m); int u, v, e, a, b, lca;
	for(int i = 1; i < n; ++i) {
		read(u); read(v); read(e);
		ins(u, v, e); ins(v, u, e);
	}
	
	_(1);
	cnt = 0; top[1] = 1;
	__(1);
	Build(1, 1, n);
	
	LL tmp;
	for(; m; --m) {
		read(e);
		if (e == 1) {
			read(u); read(v); read(a); read(b);
			lca = LCA(u, v);
			tmp = 1LL * a * di[u] + b;
			for(; top[u] != top[lca]; u = fa[top[u]])
				inc(1, 1, n, pos[top[u]], pos[u], - a, tmp);
			inc(1, 1, n, pos[lca], pos[u], - a, tmp);
			tmp -= (di[lca] << 1) * a;
			for(; top[v] != top[lca]; v = fa[top[v]])
				inc(1, 1, n, pos[top[v]], pos[v], a, tmp);
			inc(1, 1, n, pos[lca], pos[v], a, tmp);
		} else {
			ans = inf; read(u); read(v);
			for(; top[u] != top[v]; u = fa[top[u]]) {
				if (di[top[u]] < di[top[v]]) swap(u, v);
				Q(1, 1, n, pos[top[u]], pos[u]);
			}
			if (di[u] > di[v]) swap(u, v);
			Q(1, 1, n, pos[u], pos[v]);
			printf("%lld\n", ans);
		}
	}
	return 0;
}

Round2加油！