Travel 并查集
题意:给一个图,若干询问,每次询问只经过边权<=w的边,x能到达的点数
并查集啊,对询问和边排序,直接合并,维护size,查询
#include<cstdio> #include<cstring> #include<iostream> #include<algorithm> #include<cmath> #define N 105000 using namespace std; int fa[N],size[N],n,m,k; struct data{ int u,v,w; }d[4*N]; bool cmpd(data a,data b){return a.w<b.w;} struct query{ int u,w,ans,id; }q[N]; bool cmpq(query a,query b){return a.w<b.w;} bool back(query a,query b){return a.id<b.id;} int find(int x){ if(fa[x]==x)return x; fa[x]=find(fa[x]); return fa[x]; } void hb(int x,int y){ x=find(x);y=find(y); if(x==y)return; fa[y]=x;size[x]+=size[y]; } int main(){ scanf("%d%d%d",&n,&m,&k); for(int i=1;i<=n;i++){fa[i]=i;size[i]=1;} for(int i=1;i<=m;i++) scanf("%d%d%d",&d[i].u,&d[i].v,&d[i].w); sort(d+1,d+m+1,cmpd); for(int i=1;i<=k;i++){ scanf("%d%d",&q[i].u,&q[i].w); q[i].id=i; } sort(q+1,q+k+1,cmpq); q[k+1].w=q[k].w+1; int now=q[1].w,ee=1,pos=1; while(now<=q[k].w){ for(;ee<=m&&d[ee].w<=now;ee++) hb(d[ee].u,d[ee].v); for(;q[pos].w==now;pos++) q[pos].ans=size[find(q[pos].u)]; now=q[pos].w; } sort(q+1,q+k+1,back); for(int i=1;i<=k;i++) printf("%d\n",q[i].ans); return 0; }
人生如梦亦如幻 朝如晨露暮如霞。