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 105000using 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;}