BZOJ1787 [Ahoi2008]Meet 紧急集合 【LCA】

1787: [Ahoi2008]Meet 紧急集合

Time Limit: 20 Sec  Memory Limit: 162 MB
Submit: 3578  Solved: 1635
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Description

Input

Output

Sample Input

6 4
1 2
2 3
2 4
4 5
5 6
4 5 6
6 3 1
2 4 4
6 6 6

Sample Output


5 2
2 5
4 1
6 0

HINT



点很少,只有三个,画一下图,手动模拟一下就会发现:我们要求的就是画一条线将三个点连起来,使线最短

由于树路径的唯一性,连接三个点的路径是唯一的,但是由于走法的不同会导致因为重复部分路径而使路径增长


未使路径最短,我们就不能走复路,所以我们只需求出三者最大的LCA,再将另外一个点走上LCA所在路径就好了

如图:


具体LCA用倍增实现,复杂度O(mlogn + nlogn),一个是预处理,一个是询问

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = head[u]; k != -1; k = edge[k].next)
using namespace std;
const int maxn = 500005,maxm = 1000005,INF = 1000000000;
inline int RD(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57) {if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57) {out = (out << 1) + (out << 3) + c - '0'; c = getchar();}
	return out * flag;
}
int dep[maxn],fa[maxn][20];
int N,M,head[maxn],nedge = 0,p[4];
struct node{int lca,u,v;}e[4];
struct EDGE{int to,next;}edge[maxm];
inline void build(int u,int v){
	edge[nedge] = (EDGE){v,head[u]}; head[u] = nedge++;
	edge[nedge] = (EDGE){u,head[v]}; head[v] = nedge++;
}
void dfs(int u,int f,int d){
	dep[u] = ++d; fa[u][0] = f;
	Redge(u) if (edge[k].to != f) dfs(edge[k].to,u,d);
}
void init(){REP(j,19) REP(i,N) fa[i][j] = fa[fa[i][j - 1]][j - 1];}
int LCA(int u,int v){
	if (dep[u] < dep[v]) u ^= v ^= u ^= v;
	int d = dep[u] - dep[v];
	for (int i = 0; (1 << i) <= d; i++)
		if ((1 << i) & d) u = fa[u][i];
	if (u == v) return u;
	for (int i = 19; i >= 0; i--)
		if (fa[u][i] != fa[v][i]) u = fa[u][i],v = fa[v][i];
	return fa[u][0];
}
void B_Sort(){
	for (int i = 1; i <= 3; i++)
		for (int j = i + 1; j <= 3; j++)
			if (dep[e[i].lca] > dep[e[j].lca]) swap(e[i],e[j]);
}
void solve(){
	int ans,u;
	while (M--){
		REP(i,3) p[i] = RD();
		for (int i = 1,k = 0; i < 3; i++)
			for (int j = i + 1; j <= 3; j++)
				e[++k].u = i,e[k].v = j,e[k].lca = LCA(p[i],p[j]);
		B_Sort();
		//REP(i,3) printf("%d and %d  lca: %d\n",e[i].u,e[i].v,e[i].lca);
		REP(i,3) if (i != e[1].u && i != e[1].v) {u = p[i];break;}
		ans = dep[p[e[1].u]] + dep[p[e[1].v]] + dep[u] - 2 * dep[e[1].lca] - dep[e[3].lca];
		printf("%d %d\n",e[3].lca,ans);
	}
}
int main(){
	memset(head,-1,sizeof(head));
	N = RD(); M = RD();
	REP(i,N - 1) build(RD(),RD());
	dfs(1,0,0); init();
	solve();
	return 0;
}



posted @ 2017-12-09 13:40  Mychael  阅读(123)  评论(0编辑  收藏  举报