P1967 货车运输

题面

思路:最大生成树+树剖+线段树维护区间最小值。

其中因为我们线段树维护边的信息很难搞我也不会qwq,所以我们把边权转化为点权。

对于一条边的权值,我们把他下放到连接的两个点中深度大的点即儿子节点中,因为如果上放权值一个点可能对应好几条出边,而每个点入边只有一条。

最后处理细节就好啦qwq。

代码:

#include <bits/stdc++.h>
using namespace std;

template<typename temp> void read(temp &x){
	x = 0; temp f = 1; char ch;
	while(!isdigit(ch = getchar())) (ch == '-') and (f = -1);
	for(x = ch^48; isdigit(ch = getchar()); x = (x<<3)+(x<<1)+(ch^48));
	x *= f;
}
template<typename temp, typename ...res> void read(temp &a, res& ...some){read(a), read(some...);}

const int maxn = 1e5;

int n, m, q, num, fa[maxn], a[maxn];
pair<pair<int,int>, int> e[maxn];

vector<pair<int,int> >v[maxn];
vector<pair<pair<int,int>, int> >qwq;

struct poutree{
	#define ls (now << 1)
	#define rs (now<<1|1)
	#define mid ((l+r)>>1)
	int cnt, fa[maxn], dep[maxn], height_son[maxn], top[maxn], dfn[maxn], id[maxn], size[maxn];
	struct no{
		int l, r, minnum;
	}node[maxn<<2];
	void build_poutree(int now){
		size[now] = 1;
		for(int i = 0; i < v[now].size(); i ++){
			int to = v[now][i].first;
			if(dep[to]) continue;
			dep[to] = dep[fa[to] = now] + 1;
			build_poutree(to);
			size[now] += size[to];
			if(size[to] > size[height_son[now]]) height_son[now] = to;
		}
	}
	void dfs(int now, int topfa){
		top[id[dfn[now]=++cnt]=now] = topfa;
		if(height_son[now]) dfs(height_son[now],topfa);
		for(int i = 0; i < v[now].size(); i ++){
			int to = v[now][i].first;
			if(fa[now] == to or to == height_son[now]) continue;
			dfs(to,to);
		}
	}
	inline void up(int now){return(void)(node[now].minnum = min(node[ls].minnum, node[rs].minnum));}
	void build_segtree(int l, int r, int now){
		node[now].l = l, node[now].r = r;
		if(l == r) return(void)(node[now].minnum = a[l]);
		build_segtree(l, mid, ls), build_segtree(mid+1, r, rs);
		return up(now);
	}
	void quary(int l, int r, int now, int &ans){
		if(node[now].l > r or node[now].r < l) return;
		if(l <= node[now].l and node[now].r <= r) return(void)(ans = min(ans, node[now].minnum));
		quary(l, r, ls, ans), quary(l, r, rs, ans);
		return up(now);
	}
	int lca(int x, int y){
		int tmp = 1<<30;
		while(top[x] != top[y]){
			if(dep[top[x]] < dep[top[y]]) swap(x,y);
			quary(dfn[top[x]], dfn[x], 1, tmp);
			x = fa[top[x]];
		}
		if(dep[x] < dep[y]) swap(x,y);
		quary(dfn[height_son[y]], dfn[x], 1, tmp);//这里注意一下是height_son[y]不是y
		return tmp;
	}
	void build(){
		for(int i = 1; i <= n; i ++){
			if(!size[i]) build_poutree(i*(dep[i]=1)), dfs(i,i);
		}
		for(int i = 1; i <= n; i ++) a[i] = 1<<29;
		for(int i = 0; i < qwq.size(); i ++){
			a[dfn[dep[qwq[i].first.first]<dep[qwq[i].first.second]?qwq[i].first.second:qwq[i].first.first]] = qwq[i].second;
                        //权值下放到深度低的点上
		}
		return build_segtree(1,n,1);
	}
}tree;

inline void add_edge(int x, int y, int z){
	v[x].push_back(make_pair(y,z));
	v[y].push_back(make_pair(x,z));
	return;
} 
inline int find(int x){
	if(fa[x] == x) return x;
	return fa[x] = find(fa[x]);
}
bool tmp(pair<pair<int,int>, int> a, pair<pair<int,int>, int> b){
	return a.second > b.second;
}

signed main(){
	read(n, m);
	for(int i = 1, x, y, z; i <= m; i ++){
		read(x,y,z);
		e[i].first.first = x, e[i].first.second = y, e[i].second = z;
	}
	sort(e+1,e+m+1,tmp);
	for(int i = 1; i <= n; i ++) fa[i] = i;
	for(int i = 1; i <= m; i ++){
		int fx = find(e[i].first.first), fy = find(e[i].first.second);
		if(fx == fy) continue;
		qwq.push_back(e[i]);
		fa[fx] = fy;
		add_edge(e[i].first.first, e[i].first.second, e[i].second);	
	}//最大生成树
	tree.build();//构树
	read(q);
	for(int x, y; q; q --){
		read(x,y);
		if(find(x) != find(y)){
			printf("-1\n");
			continue;
		}else printf("%d\n", tree.lca(x,y));
	}
	return 0;
}
posted @ 2020-07-28 21:22  Vanyun  阅读(83)  评论(2编辑  收藏  举报