poj 3237 tree

Tree

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 7409 Accepted: 2009

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 vChange the weight of the ith edge to vNEGATE a bNegate the weight of every edge on the path from a to bQUERY a bFind 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 bwith 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

131 2 12 3 2QUERY 1 2CHANGE 1 3QUERY 1 2DONE

Sample Output

13

Source

POJ Monthly--2007.06.03, Lei, Tao

题目大意： 有T组数据，读入节点数N，和n-1 条边的信息，接着给出多次操作，操作以DONE结束。

QUERY  X,Y 表示求x到y 路径上的最大边权

CHANGE  X Y 表示把第X条路径的值改为Y

NEGATE  X Y 表示把X 到Y 路径上的值全部取相反数

题解： 把边权下放为点权

用线段树记录区间最小值和最大值，每次取反时，只需令max=-min, min=-max即可

处理时要格外注意细节！！  

因为边权下放为点权，所以求X,Y的路径是不能计算X点

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#define N 10003#define M 20003using namespace std;int n,m,point[N],next[M],v[M],c[M],deep[N],size[N],pos[N],val[N],tv[N];int tr[40003],tr1[40003],delta[40003],tot,sz,t,fa[N][15],mi[15],son[N],belong[N],ed[M],e[M];int xx,yy; bool p;const int inf=1e9;void clear(){  tot=0; sz=0;  memset(delta,0,sizeof(delta)); memset(point,0,sizeof(point));  memset(next,0,sizeof(next)); memset(tr,0,sizeof(tr)); memset(tr1,0,sizeof(tr1));  memset(fa,0,sizeof(fa)); memset(son,0,sizeof(son)); memset(e,0,sizeof(e));  memset(ed,0,sizeof(ed)); memset(deep,0,sizeof(deep));}void ne(int &x,int &y){  int t=x;  x=(-1)*y;  y=(-1)*t;}void add(int x,int y,int z,int num){  tot++; next[tot]=point[x]; point[x]=tot; v[tot]=y; c[tot]=z; e[tot]=num;  tot++; next[tot]=point[y]; point[y]=tot; v[tot]=x; c[tot]=z; e[tot]=num;}void dfs1(int x,int f,int dep){  size[x]=1; deep[x]=dep;  for (int i=1;i<=13;i++)   {   if (deep[x]-mi[i]<0) break;   fa[x][i]=fa[fa[x][i-1]][i-1];   }  int k=0;  for (int i=point[x];i!=0;i=next[i])   if (v[i]!=f)    {      fa[v[i]][0]=x;      val[v[i]]=c[i];  ed[e[i]]=v[i];//第I条边下放到了哪一个点       dfs1(v[i],x,dep+1);      if(size[v[i]]>size[k])       k=v[i];      son[x]=k;      size[x]+=size[v[i]];    }}void dfs2(int x,int chain){  pos[x]=++sz; tv[sz]=val[x]; belong[x]=chain;  if (son[x])  dfs2(son[x],chain);  for (int i=point[x];i!=0;i=next[i])   if (deep[v[i]]>deep[x]&&v[i]!=son[x])    dfs2(v[i],v[i]);}int lca(int x,int y){  if (deep[x]<deep[y])  swap(x,y);  int k=deep[x]-deep[y];  for (int i=0;i<=13;i++)   if (k>>i&1)  x=fa[x][i];  if (x==y)    return x;  for (int i=13;i>=0;i--)   if(fa[x][i]!=fa[y][i])    x=fa[x][i],y=fa[y][i];  return fa[x][0];} void pushdown(int now,int l,int r){  if (!delta[now]) return;  ne(tr[now<<1],tr1[now<<1]);  delta[now<<1]^=1;  ne(tr[now<<1|1],tr1[now<<1|1]);  delta[now<<1|1]^=1;  delta[now]=0;}void update(int x){tr[x]=max(tr[x<<1],tr[x<<1|1]);tr1[x]=min(tr1[x<<1],tr1[x<<1|1]);}void build(int now,int l,int r){  if (l==r)   {   tr[now]=tr1[now]=tv[l];   return;   }  int mid=(l+r)/2;  build(now<<1,l,mid);  build(now<<1|1,mid+1,r);  update(now);}void change(int now,int l,int r,int x,int v){  if (l==r)   {   tr[now]=tr1[now]=v;   return;   }  pushdown(now,l,r);  int mid=(l+r)/2;  if (x<=mid)   change(now<<1,l,mid,x,v);  else   change(now<<1|1,mid+1,r,x,v);  update(now);}int qjmax(int now,int l,int r,int ll,int rr){   int ans=-inf;  if (ll<=l&&r<=rr)   {     return tr[now];   }  pushdown(now,l,r);  int mid=(l+r)/2;  if (ll<=mid)   ans=max(ans,qjmax(now<<1,l,mid,ll,rr));  if (rr>mid)   ans=max(ans,qjmax(now<<1|1,mid+1,r,ll,rr));  return ans;}void qjch(int now,int l,int r,int ll,int rr){if (ll<=l&&r<=rr) { delta[now]^=1;    ne(tr[now],tr1[now]);    return; }pushdown(now,l,r);int mid=(l+r)/2;if (ll<=mid)  qjch(now<<1,l,mid,ll,rr);if (rr>mid) qjch(now<<1|1,mid+1,r,ll,rr);update(now);}int solve(int x,int f){  int maxn=-inf;  while (belong[x]!=belong[f])   {    maxn=max(maxn,qjmax(1,1,m,pos[belong[x]],pos[x]));    x=fa[belong[x]][0];   }  if (pos[f]+1<=pos[x])   maxn=max(maxn,qjmax(1,1,m,pos[f]+1,pos[x]));  return maxn;}void solve1(int x,int f){  while (belong[x]!=belong[f])   {    qjch(1,1,m,pos[belong[x]],pos[x]);    x=fa[belong[x]][0];   }  if (pos[f]+1<=pos[x])    qjch(1,1,m,pos[f]+1,pos[x]);}int main(){  scanf("%d",&t);  mi[0]=1;  for (int i=1;i<=13;i++)  mi[i]=mi[i-1]*2;  for (int i=1;i<=t;i++)   {   clear();   scanf("%d",&m);   for (int j=1;j<=m-1;j++)    {       int x,y,z; scanf("%d%d%d",&x,&y,&z);       add(x,y,z,j);    }   val[1]=-inf;   dfs1(1,0,1); dfs2(1,1);   char s[10];    build(1,1,m);   while (scanf("%s",s))    {    p=false;    if (s[0]=='D') break;    if (s[0]=='N')     {       int x,y;  scanf("%d%d",&x,&y);       int t=lca(x,y);       solve1(x,t),solve1(y,t);     }    if (s[0]=='C')     {     int x,y; scanf("%d%d",&x,&y);     val[ed[x]]=y;     change(1,1,m,pos[ed[x]],y);     }    if (s[0]=='Q')     {     int x,y; scanf("%d%d",&x,&y);     int t=lca(x,y);     printf("%d\n",max(solve(x,t),solve(y,t)));     }    }   }}