CF 463E Caisa and Tree (模拟)

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这道E题出奇的简单，直接模拟就行。

先建树，建树的同时记录每个节点的父节点，查询类型为1的时候直接从v的父节点往上找，找到了就直接返回，查询类型为2的时候就直接修改数组的值就好了。

#include<cstdio>#include<algorithm>#include<cstring>#include<cstdlib>#define M(a,b) memset(a,b,sizeof(a))#define ll long long#define rep(i,a,b) for(i=a;i<=b;i++)using namespace std;int a[100005],head[100005],e,v;int fa[100005];bool vst[100005];struct E{    int to,next;}edge[200005];inline void addedge(int u,int v){    edge[e].to=v;    edge[e].next=head[u];    head[u]=e++;    edge[e].to=u;    edge[e].next=head[v];    head[v]=e++;}int gcd(int a,int b){    if(!b) return a;    else return gcd(b,a%b);}void build_tree(int u){    vst[u]=1;    int i,k;    for(i=head[u];i;i=edge[i].next)    {        k=edge[i].to;        if(vst[k]) continue;        fa[k]=u;        build_tree(k);    }}int solve(int u){    if(!u) return -1;    if(gcd(a[u],a[v])>1) return u;    return solve(fa[u]);}int main(){    int n,q,i,x,y;    e=1;    scanf("%d%d",&n,&q);    rep(i,1,n) scanf("%d",&a[i]);    rep(i,2,n) scanf("%d%d",&x,&y),addedge(x,y);    build_tree(1);    rep(i,1,q)    {        int type;        scanf("%d",&type);        if(type==1)        {            scanf("%d",&v);            printf("%d\n",solve(fa[v]));        }        else        {            int w;            scanf("%d%d",&v,&w);            a[v]=w;        }    }    return 0;}

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