CF487 E. Tourists [点双连通分量 树链剖分 割点]

E. Tourists

题意:

无向连通图

  • C a w: 表示 a 城市的纪念品售价变成 w。
  • A a b: 表示有一个游客要从 a 城市到 b 城市,你要回答在所有他的旅行路径中最低售价的最低可能值。

\(1≤n,m,q≤10^5,1≤w_i\le10^9\)


显然一个点双连通分量中想去任何点都是可以的。

那么bcc缩点,树剖一下就好了?

割点可以存在于多个bcc!

所以把割点单独拿出来,向每个bcc连边


修改割点的权值怎么办?

每个割点的信息合并到父亲bcc里,查询的时候lca为bcc那么额外加上父亲割点的信息就行了


同一个bcc以及孩子割点还要用个multiset维护才能字词修改...


然后吐槽一下bcc,行为真的很奇怪啊,甚至\(u—v\)这样的东西都会当做一个bcc...

写+调了4个多小时…6.5kb...到半夜12:15,然后今天11点才起床没去上学2333

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define fir first
#define sec second
const int N = 2e5+5, inf = 1e9+5;
inline int read() {
    char c=getchar(); int x=0,f=1;
    while(c<'0'||c>'9') {if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9') {x=x*10+c-'0';c=getchar();}
    return x*f;
}

int n, m, Q, w[N], x, y, lim;
char s[N];
struct meow {int u, v;} g[N];

namespace ufs {
	int fa[N];
	int find(int x) {return x == fa[x] ? x : fa[x] = find(fa[x]);}
} using ufs::find;

struct edge {int v, ne;} e[N<<1];
int cnt, h[N];
map<int, int> has[N];
inline void ins(int u, int v) {
    if(has[u][v]) return; has[u][v] = 1;
    if(find(u) == find(v)) return;
    //printf("ins %d %d\n", u, v);
	ufs::fa[find(u)] = find(v);
	e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;
	e[++cnt] = (edge) {u, h[v]}; h[v] = cnt;
}
inline void link(int u, int v) {ins(u, v);}
namespace G {
	struct edge {int v, ne;} e[N<<1];
	int cnt, h[N];
	inline void ins(int u, int v) {
		e[++cnt] = (edge) {v, h[u]}; h[u] = cnt;
		e[++cnt] = (edge) {u, h[v]}; h[v] = cnt;
	}
	int dfn[N], low[N], dfc, is_cut[N], bel[N], bcc, bcc_size[N];
	struct meow {int u, v;} st[N]; int top;
	void dfs(int u, int fa) {
		dfn[u] = low[u] = ++dfc;
		int child = 0;
		for(int i=h[u]; i; i=e[i].ne) {
			int v = e[i].v;
			if(v == fa) continue;
			meow cur_edge = (meow){u, v};
			if(!dfn[v]) {
				st[++top] = cur_edge;
				child++;
				dfs(v, u);
				low[u] = min(low[u], low[v]);
				if(low[v] >= dfn[u]) {
					is_cut[u] = true;
					bcc++;
					while(true) {
						meow x = st[top--];
						if(bel[x.u] != bcc) {
                            bel[x.u] = bcc, bcc_size[bcc]++;
                            if(is_cut[x.u]) link(n+bcc, x.u);
                        }
						if(bel[x.v] != bcc) {
                            bel[x.v] = bcc, bcc_size[bcc]++;
                            if(is_cut[x.v]) link(n+bcc, x.v);
                        }
						if(x.u == u && x.v == v) break;
					}
				}
			} else if(dfn[v] < dfn[u]) st[++top] = cur_edge, low[u] = min(low[u], dfn[v]);
		}
		if(fa == 0 && child == 1) is_cut[u] = false;
	}
}
using G::is_cut; using G::bcc; using G::bel; using G::bcc_size;


int root, no_cut = 1;
multiset<int> q[N];

void build_tree() {

	//for(int i=1; i<=n<<1; i++) if(is_cut[i]) printf("cut %d\n", i);
    for(int i=1; i<=n; i++) bel[i] += n;// printf("bel %d  %d\n", i, bel[i]);
    using G::e; using G::h;
    for(int u=1; u<=n; u++) {
        if(is_cut[u]) {
            for(int i=h[u]; i; i=e[i].ne) {
                int v = e[i].v;
                if(!is_cut[v]) ins(bel[v], u);
            }
        } else q[bel[u]].insert(w[u]);// printf("insert %d  %d %d\n", u, bel[u], w[u]);
    }
	for(int i=1; i<=m; i++) {
		int u = g[i].u, v = g[i].v; //printf("(%d, %d)   %d  %d\n", u, v, find(u), find(v));
		if(is_cut[u] && is_cut[v] && find(u) != find(v)) ins(u, v);
	}
	if(is_cut[1]) root = 1;
	else root = bel[1];
	for(int i=1; i<=n; i++) if(is_cut[i]) no_cut = 0;
}

int dfn[N], dfc, deep[N], top[N], fa[N], size[N], mx[N];
void dfs(int u) {
	size[u] = 1;
	for(int i=h[u]; i; i=e[i].ne) {
		int v = e[i].v;
		if(v == fa[u]) continue;
		fa[v] = u;
		deep[v] = deep[u] + 1;
		if(!is_cut[u] && is_cut[v]) q[u].insert(w[v]);
		dfs(v);
		size[u] += size[v];
		if(size[v] > size[mx[u]]) mx[u] = v;
	}
}
int val[N];
void dfs(int u, int anc) {
	dfn[u] = ++dfc; val[dfc] = w[u];
	top[u] = anc;
	if(mx[u]) dfs(mx[u], anc);
	for(int i=h[u]; i; i=e[i].ne) {
		int v = e[i].v;
		if(v != fa[u] && v != mx[u]) dfs(v, v);
	}
}

namespace seg_t {
	#define lc x<<1
	#define rc x<<1|1
	#define mid ((l+r)>>1)
	#define lson lc, l, mid
	#define rson rc, mid+1, r
	int t[N<<2];
	void build(int x, int l, int r) {
		if(l == r) t[x] = val[l];
		else {
			build(lson);
			build(rson);
			t[x] = min(t[lc], t[rc]);
		}
	}
	void cha(int x, int l, int r, int p, int v) {
		if(l == r) t[x] = v;
		else {
			if(p <= mid) cha(lson, p, v);
			else cha(rson, p, v);
			t[x] = min(t[lc], t[rc]);
		}
	}
	int que(int x, int l, int r, int ql, int qr) {
		if(ql <= l && r <= qr) return t[x];
		else {
			int ans = inf;
			if(ql <= mid) ans = min(ans, que(lson, ql, qr));
			if(mid < qr)  ans = min(ans, que(rson, ql, qr));
			return ans;
		}
	}
} using seg_t::cha; using seg_t::que;

void bcc_cha(int id, int u, int ww) {
	q[id].erase(q[id].find(w[u]));
	w[u] = ww;
	q[id].insert(w[u]);
	w[id] = *q[id].begin();
	if(!no_cut) cha(1, 1, dfc, dfn[id], w[id]);
}
void bcc_cha2(int id, int last_w, int now_w) { //printf("bbc_cha2 %d  %d %d\n", id, last_w, now_w);
	q[id].erase(q[id].find(last_w));
	q[id].insert(now_w);
    //for(set<int>::iterator it = q[id].begin(); it != q[id].end(); it++) printf("%d ", *it); printf("  q\n");
	w[id] = *q[id].begin();
	cha(1, 1, dfc, dfn[id], w[id]);
}

void change(int u, int ww) {
	if(is_cut[u]) {
		int last_w = w[u];
		w[u] = ww, cha(1, 1, dfc, dfn[u], ww);
		int f = fa[u]; //printf("f %d\n", f);
		if(f && !is_cut[f]) bcc_cha2(f, last_w, ww);
	} else bcc_cha(bel[u], u, ww);
}

void query(int x, int y) {
	if(x == y) {printf("%d\n", w[x]); return;}
	if(!is_cut[x]) x = bel[x];
	if(!is_cut[y]) y = bel[y];
	//printf("query %d %d\n", x, y);
	int ans = inf;
	while(top[x] != top[y]) {
		if(deep[top[x]] < deep[top[y]]) swap(x, y);
		ans = min(ans, que(1, 1, dfc, dfn[top[x]], dfn[x]));
		x = fa[top[x]];
	}
	if(dfn[x] > dfn[y]) swap(x, y);
	ans = min(ans, que(1, 1, dfc, dfn[x], dfn[y]));
	if(!is_cut[x] && fa[x]) ans = min(ans, w[fa[x]]);
	printf("%d\n", ans);
}

int main() {
	freopen("in", "r", stdin);
	n = read(); m = read(); Q = read();
	for(int i=1; i<=n; i++) w[i] = read();
	for(int i=1; i<=m; i++) g[i].u = read(), g[i].v = read(), G::ins(g[i].u, g[i].v);
    for(int i=1; i<=n<<1; i++) ufs::fa[i] = i;

	G::dfs(1, 0);
	build_tree();
	if(no_cut) {
        w[n+1] = *q[n+1].begin();
		for(int i=1; i<=Q; i++) {
			scanf("%s", s); x = read(); y = read();
			if(s[0] == 'C') bcc_cha(n+1, x, y);
			else printf("%d\n", x==y ? w[x] : w[n+1]);
		}
		return 0;
	}

	dfs(root); dfs(root, root);
    for(int i=1; i<=bcc; i++) w[i+n] = *q[i+n].begin(), val[dfn[i+n]] = w[i+n];
	seg_t::build(1, 1, dfc); //puts("hi");

	for(int i=1; i<=Q; i++) { //printf("\n--------Q----------- %d\n", i);
		scanf("%s", s); x = read(); y = read();
		if(s[0] == 'C') change(x, y);
		else query(x, y);
	}
}

posted @ 2017-05-15 12:30  Candy?  阅读(305)  评论(0编辑  收藏  举报