HDU 3078 倍增LCA+排序

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题意：n个点，n-1条边构成无向树，每个节点有权，Q次询问，每次或问从a->b的最短路中,权第k大的值，或者更新节点a的权。

分析：首先倍增LCA预处理出祖先，然后暴力查找排序即可，水题。。。

代码：

#pragma comment(linker,"/STACK:102400000,102400000")#include <iostream>#include <string.h>#include <stdio.h>#include <algorithm>#include <vector>#include <string>#include <math.h>#include <queue>#include <stack>#include <map>#include <set>using namespace std;typedef long long ll;const int M=30;const int maxn=100005;const int maxm=200005;struct EdgeNode{    int to;    int next;}edge[maxm];int head[maxn],cnt,root;void add(int x,int y){    edge[cnt].to=y;    edge[cnt].next=head[x];    head[x]=cnt++;}int fa[maxn][M],deep[maxn],zhi[maxn],s[maxn];void init(){    cnt=0;    memset(head,-1,sizeof(head));    memset(deep,-1,sizeof(deep));    memset(fa,0,sizeof(fa));}void dfs(int pre,int u) //预处理{    deep[u]=deep[pre]+1;    fa[u][0]=pre;    for(int i=1;i<M;i++)fa[u][i]=fa[fa[u][i-1]][i-1]; //预处理出祖先    for(int i=head[u];i!=-1;i=edge[i].next)    {        int v=edge[i].to;        if(v!=pre)dfs(u,v);    }}int LCA(int u,int v){    if(deep[u]<deep[v])swap(u,v);    int d=deep[u]-deep[v];    for(int i=0;i<M;i++)if((1<<i)&d)u=fa[u][i]; //走到同一层    if(u==v)return u;    for(int i=M-1;i>=0;i--)  //二分计算LCA    {        if(fa[u][i]!=fa[v][i])        {            u=fa[u][i];            v=fa[v][i];        }    }    u=fa[u][0];    return u;}void solve(int t,int x,int y,int k){    int ans=0;    s[ans++]=zhi[x]; s[ans++]=zhi[y];    while(x!=t&&y!=t)    {        if(deep[x]>deep[y]){            x=fa[x][0];            s[ans++]=zhi[x];        }        else{            y=fa[y][0];            s[ans++]=zhi[y];        }    }    sort(s,s+ans);    printf("%d\n",s[ans-k]);}int main(){    int i,x,y,k,ans,n,m;    while(~scanf("%d%d",&n,&m))    {        init();        for(i=1;i<=n;i++)scanf("%d",&zhi[i]);        for(i=1;i<n;i++){            scanf("%d%d",&x,&y);            add(x,y);            add(y,x);        }        dfs(-1,1);        for(i=1;i<=m;i++){            scanf("%d%d%d",&k,&x,&y);            if(k==0)zhi[x]=y;            else{                if(y==0||y>n){printf("invalid request!");continue;}                int t=LCA(x,y);                if(deep[x]+deep[y]-2*deep[t]+1<k)printf("invalid request!\n");                else{                    solve(t,x,y,k);                }            }        }    }    return 0;}

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