【poj3237】【Tree】

1583. [POJ3237]树的维护

★★★☆ 输入文件:maintaintree.in 输出文件:maintaintree.out 简单对比

时间限制:5 s 内存限制:128 MB

【题目描述】

给你由N个结点组成的树。树的节点被编号为1到N,边被编号为1到N-1。每一条边有一个权值。然后你要在树上执行一系列指令。指令可以是如下三种之一:

CHANGE i v:将第i条边的权值改成v。

NEGATE a b:将点a到点b路径上所有边的权值变成其相反数。

QUERY a b:找出点a到点b路径上各边的最大权值。

【输入格式】

输入文件的第一行有一个整数N(N<=10000)。

接下来N-1行每行有三个整数a,b,c,代表点a和点b之间有一条权值为c的边。这些边按照其编号从小到大给出。

接下来是若干条指令(不超过10^5条),都按照上面所说的格式。

输入文件的最后一行是”DONE”.

【输出格式】

对每个“QUERY”指令,输出一行,即路径上各边的最大权值。

【样例输入】

3

1 2 1

2 3 2

QUERY 1 2

CHANGE 1 3

QUERY 1 2

DONE

【样例输出】

1

3

思路:这道题只是比平常的多了一个置反操作,我们止血药打一个标记就好了。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int N=10100;
int n,point[N],next[N*10],tot=1,siz[N]={0},fa[N][20]={0},son[N]={0};
int deep[N]={0},num=0,pos[N]={0},belong[N]={0},r[N]={0},map[N]={0},de[N*4]={0};
struct S{
    int st,en,va;
}aa[N*10];
struct C{
    int maxn,minn,lazy;
}tr[N*4];
bool use[N];
inline void add(int x,int y,int z)
{
    tot+=1;next[tot]=point[x];point[x]=tot;
    aa[tot].st=x;aa[tot].en=y;aa[tot].va=z;
    tot+=1;next[tot]=point[y];point[y]=tot;
    aa[tot].st=y;aa[tot].en=x;aa[tot].va=z;
}
inline void dfs_1(int x)
{
    int i;
    siz[x]=1;
    use[x]=false;
    for(i=1;i<=14;++i){
        if(deep[x]<(1<<i)) break;
        fa[x][i]=fa[fa[x][i-1]][i-1];
    }
    for(i=point[x];i;i=next[i])
      if(use[aa[i].en]){
        fa[aa[i].en][0]=x;
        deep[aa[i].en]=deep[x]+1;
        dfs_1(aa[i].en);
        siz[x]+=siz[aa[i].en];
      }
}
inline void dfs_2(int x,int y)
{
    int i,k=0;
    num+=1;
    belong[x]=y;
    for(i=point[x];i;i=next[i]){
        if(deep[aa[i].st]>deep[aa[i].en]) map[aa[i].st]=i>>1;
        else map[aa[i].en]=i>>1;
    }
    for(i=point[x];i;i=next[i])
      if(deep[x]<deep[aa[i].en]&&siz[k]<siz[aa[i].en])
        k=aa[i].en,son[x]=i>>1,pos[i>>1]=num;
    if(k==0) {num-=1; return ;}
    dfs_2(k,y);
    for(i=point[x];i;i=next[i])
      if(deep[x]<deep[aa[i].en]&&k!=aa[i].en)
        num+=1,pos[i>>1]=num,dfs_2(aa[i].en,aa[i].en);
}
inline int LCA(int x,int y)
{
    int i;
    if(deep[x]<deep[y])swap(x,y);
    int t=deep[x]-deep[y];
    for(i=0;i<=14;++i)
        if(t&(1<<i))x=fa[x][i];
    for(i=14;i>=0;--i)
        if(fa[x][i]!=fa[y][i])
        {x=fa[x][i];y=fa[y][i];}
    if(x==y)return x;
    else return fa[x][0];
}
#define mid (l+r)/2
#define L k<<1,l,mid
#define R k<<1|1,mid+1,r
inline void solve(int k)
{
    swap(tr[k].maxn,tr[k].minn);
    tr[k].maxn=-tr[k].maxn;
    tr[k].minn=-tr[k].minn;
}
inline void pushdown(int k)
{
    if(de[k]!=0){
        if((k<<1)<N*4){
            solve(k<<1); de[k<<1]=!de[k<<1];
            solve(k<<1|1); de[k<<1|1]=!de[k<<1|1];
        }
        de[k]=0;
    }
}
inline void insert(int k,int l,int r,int x,int y)
{
    pushdown(k);
    if(l==r&&l==x){
        tr[k].maxn=tr[k].minn=y;
        return ;
    }
    if(x<=mid) insert(L,x,y);
    else insert(R,x,y);
    tr[k].maxn=max(tr[k<<1].maxn,tr[k<<1|1].maxn);
    tr[k].minn=min(tr[k<<1].minn,tr[k<<1|1].minn);
}
inline int qurry(int k,int l,int r,int x,int y)
{
    int maxn=-210000000;
    pushdown(k);
    if(x<=l&&y>=r) return tr[k].maxn;
    if(x<=mid) maxn=max(maxn,qurry(L,x,y));
    if(y>mid) maxn=max(maxn,qurry(R,x,y));
    return maxn;
}
inline int ask(int x,int y)
{
    int maxn=-210000000,z;
    if(map[x]==0) return maxn;
    if(belong[x]==x&&x!=y){
        maxn=max(maxn,qurry(1,1,n-1,pos[map[x]],pos[map[x]]));
        x=fa[x][0];
    }
    if(map[x]==0) return maxn;
    while(belong[x]!=belong[y]){
        maxn=max(maxn,qurry(1,1,n-1,pos[son[belong[x]]],pos[map[x]]));
        z=fa[belong[x]][0];
        maxn=max(maxn,qurry(1,1,n-1,pos[map[belong[x]]],pos[map[belong[x]]]));
        x=z;
    }
    if(map[x]==0) return maxn;
    maxn=max(maxn,qurry(1,1,n-1,pos[son[y]],pos[map[x]]));
    return maxn;
}
inline void change(int k,int l,int r,int x,int y)
{
    pushdown(k);
    if(x<=l&&y>=r){
        solve(k); de[k]=1;
        return ;
    }
    if(x<=mid) change(L,x,y);
    if(y>mid) change(R,x,y);
    tr[k].maxn=max(tr[k<<1].maxn,tr[k<<1|1].maxn);
    tr[k].minn=min(tr[k<<1].minn,tr[k<<1|1].minn);
}
inline void negate_(int x,int y)
{
    int z;
    if(map[x]==0) return ;
    if(belong[x]==x&&x!=y){
        change(1,1,n-1,pos[map[x]],pos[map[x]]);
        x=fa[x][0];
    }
    if(map[x]==0) return ;
    while(belong[x]!=belong[y]){
        change(1,1,n-1,pos[son[belong[x]]],pos[map[x]]);
        z=fa[belong[x]][0];
        change(1,1,n-1,pos[map[belong[x]]],pos[map[belong[x]]]);
        x=z;
    }
    if(map[x]==0) return ;
    change(1,1,n-1,pos[son[y]],pos[map[x]]);
    return ;
}
int main()
{
    int i,j,x,y,z;
    char ch[10];
    memset(use,1,sizeof(use));
    scanf("%d",&n);
    for(i=1;i<n;++i){
        scanf("%d%d%d",&x,&y,&z);
        add(x,y,z);r[i]=z;
    }
    dfs_1(1);
    dfs_2(1,1);
    for(i=1;i<n;++i) insert(1,1,n-1,pos[i],r[i]);
    while(scanf("%*c%s",&ch)){
        if(ch[0]==‘D‘) break;
        scanf("%d%d",&x,&y);
        if(ch[0]==‘C‘) insert(1,1,n-1,pos[x],y);
        if(ch[0]==‘Q‘){
            int lca=LCA(x,y);
            printf("%d\n",max(ask(x,lca),ask(y,lca)));
        }
        if(ch[0]==‘N‘){
            int lca=LCA(x,y);
            negate_(x,lca);negate_(y,lca);
        }
    }
}
时间: 2024-10-13 22:49:40

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