[POJ-3237] [Problem E]

Tree

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 13156		Accepted: 3358

题目链接

http://poj.org/problem?id=3237

Description

You are given a tree with N nodes. The tree’s nodes are numbered 1 through N and its edges are numbered 1 through N − 1. Each edge is associated with a weight. Then you are to execute a series of instructions on the tree. The instructions can be one of the following forms:

CHANGE i v	Change the weight of the ith edge to v
NEGATE a b	Negate the weight of every edge on the path from a to b
QUERY a b	Find the maximum weight of edges on the path from a to b

Input

The input contains multiple test cases. The first line of input contains an integer t (t ≤ 20), the number of test cases. Then follow the test cases.

Each test case is preceded by an empty line. The first nonempty line of its contains N (N ≤ 10,000). The next N − 1 lines each contains three integers a, b and c, describing an edge connecting nodes a and b with weight c. The edges are numbered in the order they appear in the input. Below them are the instructions, each sticking to the specification above. A lines with the word “DONE” ends the test case.

Output

For each “QUERY” instruction, output the result on a separate line.

Sample Input

 

1

​

3

1 2 1

2 3 2

QUERY 1 2

CHANGE 1 3

QUERY 1 2

DONE

Sample Output

 

1

3

题解

树练剖分的模板题，单点修改，区间查询，还有区间取反（就是变成相反数。。。）。

注意一下单点的lazy标记处理就好了，我是每个区间权值由子区间和lazy标记组成，

然后叶子节点lazy区间用一次就消失。

代码

 #include<cstdio>
 #include<iostream>
 #include<algorithm>
 #include<cstring>
 using namespace std;
 #define ll long long
 #define N 100050
 #define INF 0x7f7f7f7f
 int n,bh[N],w[N];
 struct Tree{int l,r,lazy,max,min;}tr[N<<];
 struct Edge{int from,to,val,id,s;}edges[N<<];
 int tot,last[N];
 int cnt,fa[N],size[N],dp[N],son[N],rk[N],kth[N],top[N];

 template<typename T>void read(T&x)
 {
   ll k=; char c=getchar();
   x=;
   while(!isdigit(c)&&c!=EOF)k^=c=='-',c=getchar();
   if (c==EOF)exit();
   while(isdigit(c))x=x*+c-'',c=getchar();
   x=k?-x:x;
 }
 void AddEdge(int x,int y,int z,int id)
 {
   edges[++tot]=Edge{x,y,z,id,last[x]};
   last[x]=tot;
 }
 void read_char(char &c)
 {while(!isalpha(c=getchar())&&c!=EOF);}
 void push_up(int x)
 {
   int len=(tr[x].r-tr[x].l+);
   if (len>)
     {
       tr[x].max=max(tr[x<<].max,tr[x<<|].max);
       tr[x].min=min(tr[x<<].min,tr[x<<|].min);
     }
   if (tr[x].lazy==-)
     {
       tr[x].max*=-;
       tr[x].min*=-;
       swap(tr[x].max,tr[x].min);
     }
   if (len==)tr[x].lazy=;
 }
 void push_down(int x)
 {
   tr[x<<].lazy*=tr[x].lazy;
   tr[x<<|].lazy*=tr[x].lazy;
   push_up(x<<);
   push_up(x<<|);
   tr[x].lazy=;
 }
 void bt(int x,int l,int r)
 {
   tr[x].l=l; tr[x].r=r; tr[x].lazy=;
   if (l==r)
     {
       tr[x].max=tr[x].min=w[kth[l]];
       return;
     }
   int mid=(l+r)>>;
   bt(x<<,l,mid);
   bt(x<<|,mid+,r);
   push_up(x);
 }
 void update(int x,int p,int tt)
 {
   if (p<=tr[x].l&&tr[x].r<=p)
     {
       tr[x].max=tr[x].min=tt;
       tr[x].lazy=;
       return;
     }
   int mid=(tr[x].l+tr[x].r)>>;
   push_down(x);
   if (p<=mid)update(x<<,p,tt);
   if (mid<p)update(x<<|,p,tt);
   push_up(x);
 }
 int neg(int x,int l,int r,int tt)
 {
   if (l<=tr[x].l&&tr[x].r<=r)
     {
       tr[x].lazy*=tt;
       push_up(x);
       return tr[x].max;
     }
   int mid=(tr[x].l+tr[x].r)>>,ans=-INF;
   push_down(x);
   if (l<=mid)ans=max(ans,neg(x<<,l,r,tt));
   if (mid<r)ans=max(ans,neg(x<<|,l,r,tt));
   push_up(x);
   return ans;
 }
 void dfs1(int x,int pre)
 {
   fa[x]=pre;
   dp[x]=dp[pre]+;
   size[x]=;
   son[x]=;
   for(int i=last[x];i;i=edges[i].s)
     {
       Edge &e=edges[i];
       if (e.to==pre)continue;
       w[e.to]=e.val;
       bh[e.id]=e.to;
       dfs1(e.to,x);
       size[x]+=size[e.to];
       if (size[e.to]>size[son[x]])son[x]=e.to;
     }
 }
 void dfs2(int x,int y)
 {
   rk[x]=++cnt;
   kth[cnt]=x;
   top[x]=y;
   if (son[x]==)return;
   dfs2(son[x],y);
   for(int i=last[x];i;i=edges[i].s)
     {
       Edge &e=edges[i];
       if (e.to==fa[x]||e.to==son[x])continue;
       dfs2(e.to,e.to);
     }
 }
 int get_max(int x,int y,int tt)
 {
   int fx=top[x],fy=top[y],ans=-INF;
   while(fx!=fy)
     {
       if (dp[fx]<dp[fy])swap(x,y),swap(fx,fy);
       ans=max(ans,neg(,rk[fx],rk[x],tt));
       x=fa[fx];fx=top[x];
     }
   if (dp[x]<dp[y])swap(x,y);
   ans=max(ans,neg(,rk[y]+,rk[x],tt));
   return ans;
 }
 void work()
 {
   read(n);
   for(int i=;i<=n-;i++)
     {
       int x,y,z;
       read(x); read(y); read(z);
       AddEdge(x,y,z,i);
       AddEdge(y,x,z,i);
     }
   dfs1(,);
   dfs2(,);
   bt(,,n);
   while()
     {
       char id; int x,y;
       read_char(id);
       if (id=='D') break;
       read(x); read(y);
       if (id=='C') update(,rk[bh[x]],y);
       if (id=='N') get_max(x,y,-);
       if (id=='Q') printf("%d\n",get_max(x,y,));
     }
 }
 void clear()
   {
     cnt=; tot=;
     memset(last,,sizeof(last));
   }
 int main()
 {
 #ifndef ONLINE_JUDGE
   freopen("aa.in","r",stdin);
 #endif
   int q;
   read(q);
   while(q--)
     {
       clear();
       work();
     }
 }