BZOJ 1015 星球大战(并查集)
正着不好搞,考虑倒着搞。倒着搞就是一个并查集。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000007 # define INF (LL)1<<60 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int res=0, flag=0; char ch; if((ch=getchar())=='-') flag=1; else if(ch>='0'&&ch<='9') res=ch-'0'; while((ch=getchar())>='0'&&ch<='9') res=res*10+(ch-'0'); return flag?-res:res; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=400005; //Code begin... int edge[N][2], node[N], fa[N], res[N]; bool vis[N]; VI head[N]; int find(int x) { int s, temp; for (s=x; fa[s]>=0; s=fa[s]) ; while (s!=x) temp=fa[x], fa[x]=s, x=temp; return s; } void union_set(int x, int y) { int temp = fa[x]+fa[y]; if (fa[x]>fa[y]) fa[x]=y, fa[y]=temp; else fa[y]=x, fa[x] = temp; } int main () { int n, m, p; mem(fa,-1); scanf("%d%d",&n,&m); FOR(i,1,m) scanf("%d%d",&edge[i][0],&edge[i][1]); scanf("%d",&p); int ans=n-p; FOR(i,1,p) scanf("%d",node+i), vis[node[i]]=1; FOR(i,1,m) { if (vis[edge[i][0]]) head[edge[i][0]].pb(edge[i][1]); if (vis[edge[i][1]]) head[edge[i][1]].pb(edge[i][0]); if (!vis[edge[i][0]] && !vis[edge[i][1]]) { int u=find(edge[i][0]), v=find(edge[i][1]); if (u!=v) union_set(u,v), --ans; } } res[p+1]=ans; for (int i=p; i>=1; --i) { int u=node[i]; for (int j=0; j<head[u].size(); ++j) { if (vis[head[u][j]]) continue; int v=find(u), w=find(head[u][j]); if (v!=w) union_set(v,w), --ans; } ++ans; res[i]=ans; vis[u]=0; } FOR(i,1,p+1) printf("%d\n",res[i]); return 0; }