LuoGuP2495:[SDOI2011]消耗战

Pre

开始的时候不叫 (O2) 的话 (60) 分，加了 (O2) 就 (50) 分。。。

Solution

直接虚树，本文目的在于说明易错点。

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <limits.h>
#define ll long long
using namespace std;

struct _in {
	const _in operator , (ll & a) const {
		a = 0;
		char k = getchar ();
		while (k > '9' || k < '0') k = getchar ();
		while (k >= '0' && k <= '9') {
			a = a * 10 + k - '0';
			k = getchar ();
		}
		return * this;
	}
}in;
inline void swap (int&u, int&v) {
	int c = u;
	u = v,
	v = c;
}
inline int min (int u, int v) {
	return u > v ? v : u;
}
inline ll min (ll u, ll v) {
	return u > v ? v : u;
}
const int N = 250000 + 5;
int n, dfn[N], dftot;
int m, dep[N], st[N][20];
ll sts[N][20], cnt, k[N];
bool cmp (int u, int v) {
	return dfn[u] < dfn[v];
}
struct ECH {
	int tme[N], now;
	bool rch[N];
	inline void init () {
		++now;
	}
	inline void insert (int p) {
		tme[p] = now, rch[p] = 1;
	}
	inline bool query (int p) {
		if (tme[p] != now) tme[p] = now, rch[p] = 0;
		else return rch[p];
		return 0;
	}
}rch;
struct Graph {
	int fr[N << 1], to[N << 1], h[N], tot;
	ll val[N << 1];
	inline void add (int u, int v, ll w) {
		tot++;
		fr[tot] = h[u];
		to[tot] = v;
		h[u] = tot;
		val[tot] = w;
	}
	inline void dfs (int u, int f) {
		dfn[u] = ++dftot;
		dep[u] = dep[f] + 1;
		st[u][0] = f;
		for (int i = 1; i <= 18; ++i) st[u][i] = st[ st[u][i - 1] ][i - 1];
		for (int i = 1; i <= 18; ++i) sts[u][i] = min (sts[u][i - 1], sts[ st[u][i - 1] ][i - 1]);
		for (int i = h[u]; i; i = fr[i]) {
			if (to[i] == f) continue;
			sts[ to[i] ][0] = val[i];
			dfs (to[i], u);
		}
	}
	inline int lca (int u, int v) {
		if (dep[u] < dep[v]) swap (u, v);
		for (int i = 19; i >= 0 && dep[u] != dep[v]; --i)
			if (dep[ st[u][i] ] >= dep[v]) u = st[u][i];
		for (int i = 19; i >= 0; --i)
			if (st[u][i] != st[v][i]) u = st[u][i], v = st[v][i];
		if (u == v) return u;
		else return st[u][0];
	}
	inline ll query (int u, int v) {
		ll res = INT_MAX;
		if (dep[u] < dep[v]) swap (u, v);
		for (int i = 18; i >= 0 && u != v; --i)
			if (st[u][i] != v) res = min (res, sts[u][i]), u = st[u][i];
		if (u != v) res = min (res, sts[u][0]);
		return res;
	}
}g1;
struct Graph2 {
	int stk[N], top;
	ll dp[N];
	inline void init () {
		top = 0;
	}
	inline void build () {
		for (int i = 1; i <= cnt; ++i) {
			dp[ k[i] ] = 0;
			int v = k[i];
			if (!top) { stk[++top] = v; continue; }
			else {
				int lca = g1.lca (v, stk[top]);
				if (lca == stk[top]) {
					stk[++top] = v;
					continue;
				}
				while (1) {
					if (dep[lca] > dep[ stk[top - 1] ]) {
						dp[lca] = 0;
						ll wt = g1.query(lca, stk[top]);
						if (rch.query(stk[top])) dp[lca] += wt;
						else dp[lca] += min (wt, dp[stk[top]]);
						stk[top] = lca, stk[++top] = v;
						break;
					}
					else if (dep[lca] == dep[ stk[top - 1] ]) {
						ll wt = g1.query(lca, stk[top]);
						if (rch.query(stk[top])) dp[lca] += wt;
						else dp[lca] += min (wt, dp[stk[top]]);
						stk[top] = v;
						break;
					}
					else {
						ll wt = g1.query(stk[top - 1], stk[top]);
						if (rch.query(stk[top])) dp[ stk[top - 1] ] += wt;
						else dp[ stk[top - 1] ] += min (wt, dp[ stk[top] ]);
						top--;
					}
				}
			}
		}
		while (top > 1) {
			ll wt = g1.query(stk[top], stk[top - 1]);
			if (rch.query(stk[top])) dp[ stk[top - 1] ] += wt;
			else dp[ stk[top - 1] ] += min (wt, dp[ stk[top] ]);
			--top;
		}
	}
}g2;

int main () {	
	memset (sts, 127, sizeof (sts));
	scanf ("%d", &n);
	for (int i = 1; i < n; ++i) {
		ll x, y, z;
		in, x, y, z;
		g1.add (x, y, z), g1.add (y, x, z);
	}
	g1.dfs (1, 0);
	scanf ("%d", &m);
	while (m--) {
		in, cnt;
		rch.init ();
		for (int i = 1; i <= cnt; ++i)
			{ in, k[i]; rch.insert( k[i] ); }
		k[++cnt] = 1;
		sort (k + 1, k + cnt + 1, cmp);
		g2.init ();
		g2.build ();
		printf ("%lld
", g2.dp[1]);
	}
	return 0;
}

Conclusion

注意建虚树时：
1、不能使用

memset (h, 0, sizeof h);

来初始化图，因为时间复杂度是错的。

2、(DP) 的时候最好在建树的时候 (DP)， 否则可能挂。