POJ 3237 Tree

Tree

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 3836		Accepted: 1088

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

树链剖分~

然后注意一下 有n-1条边。。每条边要addedge 两次 然后，注意一下空间开两倍就可以了。否则RE

只要记录最小最大值。

取反的时候把最小最大交换一下  ，然后在取负就可以了

还有lazy更新那个取反

pushdown的时候要注意一下把孩子的lazy取反

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;

#define root 1,n,1
#define lr rt<<1
#define rr rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1

const int inf=1e9;
const int N=20010;

struct node
{
    int u,v,w;
}e[N];


int n;
int top[N],rnk[N],fa[N],p[N],son[N],siz[N],dep[N],pos;
int eh[N],et[N],nxt[N],tot;

void init()
{
    memset(eh,-1,sizeof eh);
    memset(son,-1,sizeof son);
    pos=1;
    tot=0;
}

void addedge(int u,int v)
{
    et[tot]=v;nxt[tot]=eh[u];eh[u]=tot++;
    et[tot]=u;nxt[tot]=eh[v];eh[v]=tot++;
}

void dfs1(int u,int father,int d)
{
    fa[u]=father;
    dep[u]=d;
    siz[u]=1;
    for(int i=eh[u]; ~i ;i=nxt[i]){
        int v=et[i];
        if( v==fa[u] )continue;
        dfs1( v,u,d+1 );
　　　　　siz[u] += siz[v] ;
        if( son[u]==-1 || siz[v] > siz[ son[u] ] )
            son[u]=v;
    }
}
void dfs2(int u,int tp)
{
    top[u]=tp;
    p[u]=pos++;
    rnk[ p[u] ]=u;
    if( son[u] == -1 ) return ;
    dfs2( son[u],tp );
    for(int i=eh[u]; ~i ;i=nxt[i] ){
        int v=et[i];
        if( v==fa[u] || v==son[u] )continue;
        dfs2(v,v);
    }
}

//-----------------------------------

int d_m[N<<2],d_M[N<<2];
bool lazy[N<<2];
void build(int l,int r,int rt)
{
    d_m[rt]=d_M[rt]=lazy[rt]=0;
    if(l==r){return ;}
    int mid=(l+r)>>1;
    build(lson);
    build(rson);

}
void Up(int rt)
{
    d_m[rt]=min( d_m[lr],d_m[rr] );
    d_M[rt]=max( d_M[lr],d_M[rr] );
}
void Down(int l,int r,int rt)
{
    if(l==r)return ;
    if( lazy[rt] ) {
        swap(d_m[rr],d_M[rr]);
        d_M[rr] = -d_M[rr];
        d_m[rr] = -d_m[rr];
        lazy[rr] ^= 1;
        swap(d_m[lr],d_M[lr]);
        d_M[lr] = -d_M[lr];
        d_m[lr] = -d_m[lr];
        lazy[lr] ^= 1;
    lazy[rt]=0;
    }
}

void update(int l,int r,int rt,int x,int v)
{

    if(l==r){
        d_m[rt]=d_M[rt]=v;lazy[rt]=0;
        return ;
    }
    Down(l,r,rt);
    int mid=(l+r) >> 1;

    if(x <= mid )
        update(lson,x,v);
    else
        update(rson,x,v);

    Up(rt);
}

int query(int l,int r,int rt,int L,int R)
{
    int res = -inf;

    if(L <= l && r<=R ){
        return d_M[rt];
    }
    Down(l,r,rt);
    int mid=(l+r)>>1;
    if( L <= mid )res=max( res,query(lson,L,R) );
    if( R  > mid )res=max( res,query(rson,L,R) );
    return res;

}

void nega(int l,int r,int rt,int L,int R)
{
    if( L <= l && r<= R ){
        swap(d_m[rt],d_M[rt]);
        d_M[rt] = -d_M[rt];
        d_m[rt] = -d_m[rt];
        lazy[rt] ^= 1;
        return ;
    }
    Down(l,r,rt);
    int mid=(l+r) >> 1;
    if( L <= mid ) nega(lson,L,R);
    if( R > mid ) nega(rson,L,R);
    Up(rt);
}

int Q(int u,int v)
{
    int res = -inf;
    int f1=top[u],f2=top[v];
    while(f1 != f2){
        if(dep[f1] < dep[f2]){
            swap(f1,f2);
            swap(u,v);
        }
        res=max(res,query(root,p[f1],p[u]));
        u = fa[ f1 ];
        f1 = top[ u ];
    }
    if( u == v )return res;
    if( dep[u] > dep[v] )swap(u,v);
    return max( res , query( root,p[ son[u] ] ,p[v] ) );
}

void NE(int u,int v)
{
    int f1=top[u],f2=top[v];
    while(f1 != f2){
        if(dep[f1] < dep[f2]){
            swap(f1,f2);
            swap(u,v);
        }
        nega(root,p[f1],p[u]);
        u = fa[ f1 ];
        f1 = top[ u ];
    }
    if( u == v )return ;
    if( dep[u] > dep[v] )swap(u,v);
    nega( root,p[ son[u] ] ,p[v]) ;
}

void run()
{
    char op[10];
    int x,y,v;
    init();
    scanf("%d",&n);
    for(int i=1;i<n;++i){
        scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
        addedge( e[i].u , e[i].v );
    }
    dfs1(1,0,0);
    dfs2(1,1);
    build( root );
    for(int i=1;i < n ;++i){
        if( dep[e[i].u] > dep[ e[i].v] )swap(e[i].u,e[i].v);
        update(root,p[ e[i].v ],e[i].w);
    }
    while(scanf("%s",op)){
        if(op[0] == 'D')break;
        scanf("%d%d",&x,&y);
        if(op[0]=='C'){
            update( root, p[ e[ x ].v ]  , y );
        }
        else if(op[0]=='Q'){
            printf("%d
",Q(x,y));
        }
        else {
            NE(x,y);
        }
    }
}
int main()
{
    int _;
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif
    scanf("%d",&_);
    while(_--)run();
    return 0;
}