【LOJ2955】「NOIP2018」保卫王国

【题目链接】

点击打开链接

【思路要点】

考虑 $O(NM)$ 的暴力，对于每个询问，我们需要进行一次树形 $dp$ 。

$dp$ 的状态大致是令某个点 $i$ 取/不取，其子树内的最优权值和。

考虑优化，对于一个询问 $(x,y)$ ，将路径 $(x,y)$ 单独考虑，对于路径 $(x,y)$ 上的每个点，其不在路径上的子树内的 $dp$ 值是可以预处理的，并且，若我们将路径 $(x,y)$ 拆分为 $(x,Lca),(y,Lca)$ ，对于路径记录路径首/尾处取/不取时的最优解，可以通过倍增预处理某一段的 $dp$ 值合并后的结果。

具体细节可见代码。

时间复杂度 $O(NLogN+MLogN)$ 。

【代码】

#include<bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 5;
const int MAXLOG = 22;
typedef long long ll;
const ll INF = 1e18;
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') f = -f;
	for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
	x *= f;
}
template <typename T> void chkmin(T &x, T y) {x = min(x, y); }
struct info {ll a[2][2]; };
info cipher() {
	info ans;
	ans.a[0][0] = INF;
	ans.a[0][1] = INF;
	ans.a[1][0] = INF;
	ans.a[1][1] = INF;
	return ans;
}
info operator + (info a, info b) {
	info ans = cipher();
	for (int i = 0; i <= 1; i++)
	for (int j = 0; j <= 1; j++) {
		chkmin(ans.a[i][j], a.a[i][0] + b.a[1][j]);
		chkmin(ans.a[i][j], a.a[i][1] + b.a[0][j]);
		chkmin(ans.a[i][j], a.a[i][1] + b.a[1][j]);
	}
	return ans;
}
int n, m, depth[MAXN], father[MAXN][MAXLOG];
ll dp[MAXN][2], fp[MAXN][2], up[MAXN][2], val[MAXN];
info v[MAXN][MAXLOG];
vector <int> a[MAXN];
void getdp(int pos, int fa) {
	dp[pos][0] = 0, dp[pos][1] = val[pos];
	for (unsigned i = 0; i < a[pos].size(); i++)
		if (a[pos][i] != fa) {
			getdp(a[pos][i], pos);
			dp[pos][0] += dp[a[pos][i]][1];
			dp[pos][1] += min(dp[a[pos][i]][0], dp[a[pos][i]][1]);
		}
	for (unsigned i = 0; i < a[pos].size(); i++)
		if (a[pos][i] != fa) {
			fp[a[pos][i]][0] = dp[pos][0] - dp[a[pos][i]][1];
			fp[a[pos][i]][1] = dp[pos][1] - min(dp[a[pos][i]][0], dp[a[pos][i]][1]);
		}
}
void getup(int pos, int fa) {
	ll tmp[2] = {dp[pos][0] + up[pos][1], dp[pos][1] + min(up[pos][0], up[pos][1])};
	for (unsigned i = 0; i < a[pos].size(); i++)
		if (a[pos][i] != fa) {
			up[a[pos][i]][0] = tmp[0] - dp[a[pos][i]][1];
			up[a[pos][i]][1] = tmp[1] - min(dp[a[pos][i]][0], dp[a[pos][i]][1]);
			getup(a[pos][i], pos);
		}
}
void work(int pos, int fa) {
	father[pos][0] = fa;
	depth[pos] = depth[fa] + 1;
	for (int i = 1; i < MAXLOG; i++)
		father[pos][i] = father[father[pos][i - 1]][i - 1];
	for (unsigned i = 0; i < a[pos].size(); i++)
		if (a[pos][i] != fa) work(a[pos][i], pos);
}
int lca(int x, int y) {
	if (depth[x] < depth[y]) swap(x, y);
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (depth[father[x][i]] >= depth[y]) x = father[x][i];
	if (x == y) return x;
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (father[x][i] != father[y][i]) {
			x = father[x][i];
			y = father[y][i];
		}
	return father[x][0];
}
info formdp(int pos, int tx) {
	info ans = cipher();
	ans.a[tx][tx] = dp[pos][tx];
	return ans;
}
info getpath(int x, int f, int tx) {
	info ans = formdp(x, tx);
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (depth[father[x][i]] > depth[f]) {
			ans = ans + v[x][i];
			x = father[x][i];
		}
	return ans;
}
info Reverse(info a) {
	swap(a.a[0][1], a.a[1][0]);
	return a;
}
info getlca(int x, int y, int f) {
	ll tmp[2] = {dp[f][0] + up[f][1], dp[f][1] + min(up[f][0], up[f][1])};
	for (int i = MAXLOG - 1; i >= 0; i--) {
		if (depth[father[x][i]] > depth[f]) x = father[x][i];
		if (depth[father[y][i]] > depth[f]) y = father[y][i];
	}
	tmp[0] -= dp[x][1] + dp[y][1];
	tmp[1] -= min(dp[x][0], dp[x][1]) + min(dp[y][0], dp[y][1]);
	info ans = cipher();
	ans.a[0][0] = tmp[0];
	ans.a[1][1] = tmp[1];
	return ans;
}
info getlcb(int x, int y, int tx) {
	ll tmp[2] = {dp[x][0] + up[x][1], dp[x][1] + min(up[x][0], up[x][1])};
	for (int i = MAXLOG - 1; i >= 0; i--)
		if (depth[father[y][i]] > depth[x]) y = father[y][i];
	tmp[0] -= dp[y][1];
	tmp[1] -= min(dp[y][0], dp[y][1]);
	info ans = cipher();
	ans.a[tx][tx] = tmp[tx];
	return ans;
}
ll getans(info a) {
	ll ans = INF;
	chkmin(ans, a.a[0][0]);
	chkmin(ans, a.a[1][0]);
	chkmin(ans, a.a[0][1]);
	chkmin(ans, a.a[1][1]);
	if (ans >= INF) return -1;
	else return ans;
}
int main() {
	freopen("defense.in", "r", stdin);
	freopen("defense.out", "w", stdout);
	char s[5];
	read(n), read(m), scanf("\n%s", s + 1);
	for (int i = 1; i <= n; i++)
		read(val[i]);
	for (int i = 1; i <= n - 1; i++) {
		int x, y; read(x), read(y);
		a[x].push_back(y);
		a[y].push_back(x);
	}
	work(1, 0);
	getdp(1, 0);
	getup(1, 0);
	for (int i = 1; i <= n; i++) {
		v[i][0] = cipher();
		v[i][0].a[0][0] = fp[i][0];
		v[i][0].a[1][1] = fp[i][1];
	}
	for (int p = 1; p < MAXLOG; p++)
	for (int i = 1; i <= n; i++)
		v[i][p] = v[i][p - 1] + v[father[i][p - 1]][p - 1];
	for (int i = 1; i <= m; i++) {
		int x, tx, y, ty;
		read(x), read(tx), read(y), read(ty);
		if (depth[x] > depth[y]) {
			swap(x, y);
			swap(tx, ty);
		}
		int z = lca(x, y);
		if (x == z) {
			info ax = getlcb(x, y, tx);
			info ay = getpath(y, x, ty);
			info ans = ax + Reverse(ay);
			printf("%lld\n", getans(ans));
		} else {
			info ax = getpath(x, z, tx);
			info ay = getpath(y, z, ty);
			info az = getlca(x, y, z);
			info ans = ax + az + Reverse(ay);
			printf("%lld\n", getans(ans));
		}
	}
	return 0;
}

【LOJ2955】「NOIP2018」保卫王国

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