poj 3237 tree
Tree
| Time Limit: 5000MS | Memory Limit: 131072K | |
| Total 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 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 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
1 3 1 2 1 2 3 2 QUERY 1 2 CHANGE 1 3 QUERY 1 2 DONE
Sample Output
1 3
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 20003
using 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)));
}
}
}
}