【bzoj3237】 Ahoi2013—连通图

http://www.lydsy.com/JudgeOnline/problem.php?id=3237 (题目链接)

题意

  给出一个无向图,$Q$组询问,每次询问将原图断掉$C$条边后是否还连通。

Solution

  CDQ图分治,并查集维护。实在写不动题了,我能说我是蒯的吗T_T:http://blog.csdn.net/creationaugust/article/details/50889351

细节

  bzoj的G++版本是多久以前的了,莫名CE。。

代码

// bzoj3237
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf 2147483640
#define Pi acos(-1.0)
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout);
using namespace std;

const int maxn=200010,maxm=5000010;
int ans[maxn],del[maxn],fa[maxn],st[maxm];
int n,m,Q,tim,top;
struct data {int c,id,e[4];}q[maxn];
struct edge {int u,v;}e[maxn];

namespace Unionset {
	int find(int x) {
		if (fa[x]==x) return x;
		st[++top]=x;st[++top]=fa[x];return fa[x]=find(fa[x]);
	}
	void Union(int u,int v) {
		if (find(u)!=find(v)) {
			st[++top]=fa[v];
			st[++top]=fa[fa[v]];
			fa[fa[v]]=fa[u];
		}
	}
}
using namespace Unionset;

void solve(int l,int r) {
	int now=top,mid=(l+r)>>1,flag=1;
	if (l==r) {
		for (int i=0;i<q[l].c && flag;i++)
			if (find(e[q[l].e[i]].u)!=find(e[q[l].e[i]].v)) flag=0;
		ans[q[l].id]=flag;
		for (;now!=top;top-=2) fa[st[top-1]]=st[top];
		return;
	}
	tim++;
	for (int i=l;i<=mid;i++)
		for (int j=0;j<q[i].c;j++) del[q[i].e[j]]=tim;
	for (int i=mid+1;i<=r;i++)
		for (int j=0;j<q[i].c;j++)
			if (del[q[i].e[j]]!=tim) Union(e[q[i].e[j]].u,e[q[i].e[j]].v);
	solve(l,mid);tim++;
	for (;now!=top;top-=2) fa[st[top-1]]=st[top];
	for (int i=mid+1;i<=r;i++)
		for (int j=0;j<q[i].c;j++) del[q[i].e[j]]=tim;
	for (int i=l;i<=mid;i++)
		for (int j=0;j<q[i].c;j++)
			if (del[q[i].e[j]]!=tim) Union(e[q[i].e[j]].u,e[q[i].e[j]].v);
	solve(mid+1,r);
	for (;now!=top;top-=2) fa[st[top-1]]=st[top];
}

int main() {
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) fa[i]=i;
	for (int i=1;i<=m;i++) scanf("%d%d",&e[i].u,&e[i].v);
	scanf("%d",&Q);tim=1;
	for (int i=1;i<=Q;i++) {
		scanf("%d",&q[i].c);q[i].id=i;
		for (int j=0;j<q[i].c;j++) scanf("%d",&q[i].e[j]),del[q[i].e[j]]=tim;
	}
	for (int i=1;i<=m;i++) if (del[i]!=tim) Union(e[i].u,e[i].v);
	solve(1,Q);
	for (int i=1;i<=Q;i++) puts(ans[i] ? "Connected" : "Disconnected");
	return 0;
}

 

posted @ 2017-03-13 21:53  MashiroSky  阅读(250)  评论(0编辑  收藏  举报