Connected Components

对并查集做前后缀和。

对于并查集的前后缀中的节点i来说，如果它在前缀后所属的连通块与它在后缀中所属的连通块不同，那么说明i就是前后缀两个不同连通块的连接点，所以可以把前后缀的两个不同连通块，并为同一连通块。

之后就是求连通块个数即可。

#include <bits/stdc++.h>
using namespace std;
const int M=1e4+5,N=505;
int n,m,xx,yy,ans,q,x,y;
int u[M],v[M],l[M][N],r[M][N],f[N];int find(int x,int i)
{if (l[i][x]==x) return x;return l[i][x]=find(l[i][x],i);
}
int find2(int x,int i)
{if (r[i][x]==x) return x;return r[i][x]=find2(r[i][x],i);
}
int find3(int x)
{if (f[x]==x) return x;return f[x]=find3(f[x]);
}int main(){scanf("%d%d",&n,&m);for (register int i=1; i<=m; ++i) scanf("%d%d",&u[i],&v[i]);for (register int i=1; i<=n; ++i) l[0][i]=i;for (register int i=1; i<=m; ++i){//for (register int j=1; j<=n; ++j) l[i][j]=l[i-1][j];memcpy(l[i],l[i-1],sizeof(l[i]));xx=find(u[i],i); yy=find(v[i],i);if (xx!=yy) l[i][xx]=yy;}for (register int i=1; i<=n; ++i) r[m+1][i]=i;for (register int i=m; i>=1; --i){memcpy(r[i],r[i+1],sizeof(r[i]));xx=find2(u[i],i); yy=find2(v[i],i);if (xx!=yy) r[i][xx]=yy;}scanf("%d",&q);while (q--){scanf("%d%d",&x,&y);for (register int i=1; i<=n; ++i) f[i]=l[x-1][i];ans=0;for (register int i=1; i<=n; ++i){xx=find3(f[i]); int now=find2(r[y+1][i],y+1);yy=find3(now);if (xx!=yy) f[xx]=yy;    }for (register int i=1; i<=n; ++i) if (find3(i)==i) ans++;printf("%d\n",ans);}
return 0;
}

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Connected Components

对并查集做前后缀和。

对于并查集的前后缀中的节点i来说，如果它在前缀后所属的连通块与它在后缀中所属的连通块不同，那么说明i就是前后缀两个不同连通块的连接点，所以可以把前后缀的两个不同连通块，并为同一连通块。

之后就是求连通块个数即可。

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