CF763E Timofey and our friends animals

题目戳这里

首先题解给的是并查集的做法。这个做法很好想,但是很难码。用线段树数来维护并查集,暴力合并。

这里推荐另一个做法,可以无视\(K\)的限制。我们给每条边加个边权,这个边权为这条边左端点的值。然后我们将询问离线,按\(r\),从小到大处理。

对于当前询问\([l,r]\)我们用lct维护所有右端点\(\le r\)的边构成的最大生成树,然后用树状数组查询生成树中边权\(\ge l\)的边有几条即可。

#include<algorithm>
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstdlib>
using namespace std;

#define lowbit(a) (a&-a)
const int maxn = 200010,maxm = 500010;
int N,K,M,tot,cnt,tree[maxn],Q; bool rev[maxn];
int ch[maxn][2],fa[maxn],ma[maxn],pma[maxn],key[maxn],stack[maxn],ans[maxn];
queue <int> team;

inline void ins(int a,int b) { for (;a <= N;a += lowbit(a)) tree[a] += b; }
inline int calc(int a) { int ret = 0; for (;a;a -= lowbit(a)) ret += tree[a]; return ret; }

inline int gi()
{
	int ret = 0,f = 1; char ch;
	do ch = getchar(); while (!(ch >= '0'&&ch <= '9')&&ch != '-');
	if (ch == '-') f = -1;
	do ret = ret*10+ch-'0',ch = getchar(); while (ch >= '0'&&ch <= '9');
	return ret*f;
}

struct Node
{
	int l,r,id;
	inline void read(int i) { l = gi(),r = gi(),id = i; }
	friend inline bool operator <(const Node &a,const Node &b) { return a.r < b.r; }
}query[maxn];
struct EDGE
{
	int u,v;
	inline void read() { u = gi(),v = gi(); if (u > v) swap(u,v); }
	friend inline bool operator <(const EDGE &a,const EDGE &b) { return a.v < b.v; }
}edge[maxm];

inline bool isroot(int x) { return ch[fa[x]][0] != x&&ch[fa[x]][1] != x; }
inline void update(int x)
{
	int lc = ch[x][0],rc = ch[x][1];
	ma[x] = key[x]; pma[x] = x;
	if (lc && ma[lc] < ma[x]) ma[x] = ma[lc],pma[x] = pma[lc];
	if (rc && ma[rc] < ma[x]) ma[x] = ma[rc],pma[x] = pma[rc];
}
inline int newnode(int w)
{
	int ret;
	if (team.empty()) ret = ++cnt; else ret = team.front(),team.pop();
	key[ret] = w; fa[ret] = ch[ret][0] = ch[ret][1] = 0;
	update(ret); return ret;
}
inline void rotate(int x)
{
	int y = fa[x],z = fa[y],l = ch[y][1] == x,r = l^1;
	if (!isroot(y)) ch[z][ch[z][1] == y] = x; fa[x] = z;
	if (ch[x][r]) fa[ch[x][r]] = y; ch[y][l] = ch[x][r];
	fa[y] = x; ch[x][r] = y; update(y); update(x);
}
inline void pushdown(int x)
{
	if (rev[x])
	{
		int lc = ch[x][0],rc = ch[x][1];
		rev[x] = false; swap(ch[x][0],ch[x][1]);
		if (lc) rev[lc] ^= 1; if (rc) rev[rc] ^= 1;
	}
}
inline void splay(int x)
{
	int top = 0,i;
	for (i = x;!isroot(i);i = fa[i]) stack[++top] = i; stack[++top] = i;
	while (top) pushdown(stack[top--]);
	while (!isroot(x))
	{
		int y = fa[x],z = fa[y];
		if (!isroot(y))
		{
			if ((ch[y][0] == x)^(ch[z][0] == y)) rotate(x);
			else rotate(y);
		}
		rotate(x);
	}
}
inline int access(int x) { int t = 0; for (;x;t = x,x = fa[x]) splay(x),ch[x][1] = t,update(x); return t; }
inline int evert(int x) { int t = access(x); rev[t] ^= 1; return t; }
inline int find(int x) { int t = access(x); while (pushdown(t),ch[t][0]) t = ch[t][0]; return t; }
inline void cut(int now)
{
	evert(now); team.push(now); ins(edge[key[now]].u,-1);
	access(edge[key[now]].u); splay(now); ch[now][1] = fa[edge[key[now]].u] = 0;
	access(edge[key[now]].v); splay(now); ch[now][1] = fa[edge[key[now]].v] = 0;
}
inline void link(int w)
{
	if (find(edge[w].u) == find(edge[w].v))
		evert(edge[w].u),cut(pma[access(edge[w].v)]);
	int now = newnode(w);
	int t1 = evert(edge[w].u),t2 = evert(edge[w].v);
	fa[t1] = fa[t2] = now; ins(edge[w].u,1);
}

int main()
{
	freopen("763E.in","r",stdin);
	freopen("763E.out","w",stdout);
	cnt = N = gi(); K = gi(); M = gi();
	for (int i = 1;i <= M;++i) edge[i].read();
	for (int i = 1;i <= N;++i) key[i] = 1<<30,update(i);
	sort(edge+1,edge+M+1);
	Q = gi();
	for (int i = 1;i <= Q;++i) query[i].read(i);
	sort(query+1,query+Q+1);
	for (int i = 1,now = 1;i <= Q;++i)
	{
		for (;now <= M&&edge[now].v <= query[i].r;) link(now++);
		ans[query[i].id] = query[i].r-query[i].l+1-(calc(query[i].r)-calc(query[i].l-1));
	}
	for (int i = 1;i <= Q;++i) printf("%d\n",ans[i]);
	fclose(stdin); fclose(stdout);
	return 0;
}
posted @ 2017-03-13 23:36  lmxyy  阅读(539)  评论(0编辑  收藏