[关键字]:并查集
[题目大意]:有n个点m条边,要炸毁k个点,问按炸毁顺序炸毁每个点后有几个联通块。
//==============================================================================================
[分析]:正着做每次还要删边求联通块复杂度很高。而倒着做先吧所有点都干掉,然后并查集求出联通块个数,每次回复一个点然后合并联通块并更新当前答案,然后倒着输出答案。
[代码]:
View Code
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<algorithm>
using namespace std;
const int MAXN=410000;
struct node
{
int y,next;
}e[MAXN];
struct rec
{
int x,y;
}c[MAXN];
int first[MAXN];
int n,m,k,tot,ans;
int a[MAXN],f[MAXN],d[MAXN];
bool b[MAXN];
void Add(int x,int y)
{
++tot;
e[tot].y=y;
e[tot].next=first[x];
first[x]=tot;
}
int Find(int k)
{
if (f[k]==k) return k;
f[k]=Find(f[k]);
return f[k];
}
void Union(int x,int y,int &ans)
{
int xx=Find(x),yy=Find(y);
if (xx!=yy) f[xx]=yy,--ans;
}
void Init()
{
scanf("%d%d",&n,&m);
for (int i=0;i<n;++i) f[i]=i;
for (int i=1;i<=m;++i)
{
scanf("%d%d",&c[i].x,&c[i].y);
Add(c[i].x,c[i].y);
Add(c[i].y,c[i].x);
}
//for (int i=1;i<=m;++i) printf("%d %d\n",c[i].x,c[i].y);
scanf("%d",&k);
memset(b,0,sizeof(b));
for (int i=1;i<=k;++i)
{
scanf("%d",&a[i]);
b[a[i]]=1;
}
//for (int i=0;i<n;++i) printf("%d %d\n",i,b[i]);
}
void Solve()
{
for (int i=1;i<=m;++i)
if (!b[c[i].x] && !b[c[i].y]) Union(c[i].x,c[i].y,ans);
ans=0;
for (int i=0;i<n;++i)
if (f[i]==i && !b[i]) ++ans;
d[k+1]=ans;
for (int i=k;i>=1;--i)
{
b[a[i]]=0,++ans;
for (int j=first[a[i]];j;j=e[j].next)
if (!b[e[j].y]) Union(a[i],e[j].y,ans);
d[i]=ans;
}
for (int i=1;i<=k+1;++i) printf("%d\n",d[i]);
}
int main()
{
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
Init();
Solve();
return 0;
}