Codeforces Round #199 (Div. 2) E. Xenia and Tree （非正规解法 分情况dfs）

E. Xenia and Tree

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Xenia the programmer has a tree consisting of n nodes. We will consider the tree nodes indexed from 1 to n. We will also consider the first node to be initially painted red, and the other nodes — to be painted blue.

The distance between two tree nodes v and u is the number of edges in the shortest path between v and u.

Xenia needs to learn how to quickly execute queries of two types:

1. paint a specified blue node in red;
2. calculate which red node is the closest to the given one and print the shortest distance to the closest red node.

Your task is to write a program which will execute the described queries.

Input

The first line contains two integers n and m (2 ≤ n ≤ 10⁵, 1 ≤ m ≤ 10⁵) — the number of nodes in the tree and the number of queries. Next n - 1 lines contain the tree edges, the i-th line contains a pair of integers a_i, b_i (1 ≤ a_i, b_i ≤ n, a_i ≠ b_i) — an edge of the tree.

Next m lines contain queries. Each query is specified as a pair of integers t_i, v_i (1 ≤ t_i ≤ 2, 1 ≤ v_i ≤ n). If t_i = 1, then as a reply to the query we need to paint a blue node v_i in red. If t_i = 2, then we should reply to the query by printing the shortest distance from some red node to node v_i.

It is guaranteed that the given graph is a tree and that all queries are correct.

Output

For each second type query print the reply in a single line.

Sample test(s)

input

5 41 22 32 44 52 12 51 22 5

output

032

ps：我的做法是非正规解法，数据水了，就过了。正规解法不会，这里有正规解法的思路：点击打开链接，以后会了再贴。

思路：

分情况讨论，如果操作1多的话，就对2进行搜索，输出答案。如果操作2多的话，就对1进行搜索，用dp[ ]保存，遇到2直接输出。

代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define maxn 100005#define INF 0x3f3f3f3fusing namespace std;int n,m,ans,cnt,cnt1,cnt2;bool vis[maxn];int pp[maxn],c[maxn],dp[maxn];int x[maxn],y[maxn];struct Node{    int v,next;}edge[2*maxn];void addedge(int u,int v){    cnt++;    edge[cnt].v=v;    edge[cnt].next=pp[u];    pp[u]=cnt;}void dfs(int u,int step)   //针对操作１多的情况{    if(step>=ans) return ;    if(c[u])    {        if(ans>step) ans=step;        return ;    }    int i,j,v;    for(i=pp[u];i;i=edge[i].next)    {        v=edge[i].v;        if(!vis[v])        {            vis[v]=1;            dfs(v,step+1);        }    }}void dfs1(int u,int step)　　//针对操作２多的情况{    if(dp[u]<=step) return ;    dp[u]=step;    int i,j,v;    for(i=pp[u];i;i=edge[i].next)    {        v=edge[i].v;        if(!vis[v])        {            vis[v]=1;            dfs1(v,step+1);        }    }}void solve(){    int i,j;    if(cnt1>cnt2)    {        c[1]=1;        for(i=1;i<=m;i++)        {            if(x[i]==1) c[y[i]]=1;            else            {                ans=INF;                memset(vis,0,sizeof(vis));                vis[y[i]]=1;                dfs(y[i],0);                printf("%d\n",ans);            }        }    }    else    {        memset(dp,0x3f,sizeof(dp));        memset(vis,0,sizeof(vis));        vis[1]=1;        dfs1(1,0);        for(i=1;i<=m;i++)        {            if(x[i]==1)            {                memset(vis,0,sizeof(vis));                vis[y[i]]=1;                dfs1(y[i],0);            }            else printf("%d\n",dp[y[i]]);        }    }}int main(){    int i,j,u,v;    while(~scanf("%d%d",&n,&m))    {        cnt=0;        memset(pp,0,sizeof(pp));        memset(c,0,sizeof(c));        for(i=1;i<n;i++)        {            scanf("%d%d",&u,&v);            addedge(u,v);            addedge(v,u);        }        cnt1=cnt2=0;        for(i=1;i<=m;i++)        {            scanf("%d%d",&x[i],&y[i]);            if(x[i]==1) cnt1++;            else cnt2++;        }        solve();    }    return 0;}