SP375 QTREE - Query on a tree
题意大意
给定\(n\)个点的树，边按输入顺序编号为\(1,2,...n-1\)，要求作以下操作： CHANGE \(i\) \(t_i\) 将第\(i\)条边权值改为\(t_i\)，QUERY \(a\) \(b\) 询问从\(a\)点到\(b\)点路径上的最大边权
有多组测试数据，每组数据以DONE结尾
题目描述
You are given a tree (an acyclic undirected connected graph) with N nodes, and edges numbered 1, 2, 3...N-1.
We will ask you to perfrom some instructions of the following form:
- CHANGE i ti : change the cost of the i-th edge to ti
or - QUERY a b : ask for the maximum edge cost on the path from node a to node b
输入输出格式
输入格式：
The first line of input contains an integer t, the number of test cases (t <= 20). t test cases follow.
For each test case:
- In the first line there is an integer N (N <= 10000),
- In the next N-1 lines, the i-th line describes the i-th edge: a line with three integers a b c denotes an edge between a, b of cost c (c <= 1000000),
- The next lines contain instructions "CHANGE i ti" or "QUERY a b",
- The end of each test case is signified by the string "DONE".
There is one blank line between successive tests.
输出格式：
For each "QUERY" operation, write one integer representing its result.
输入输出样例
输入样例#1：
1
3
1 2 1
2 3 2
QUERY 1 2
CHANGE 1 3
QUERY 1 2
DONE
输出样例#1：
1
3
思路：
其实这个题也不算特别的难，如果你做过有关换边权为点权的题目的话，比如……月下“毛景树”，旅游等。自我感觉难度都比这道题要大，并且都涉及到化边权为点权然后树链剖分。
不多说了，我们来看这个题，这道题与普通的树链剖分题目不同的是，这道题目给出的是边权，但是普通的树链剖分题目都是给出的点权，那，该怎么办呢？不难发现，每个点有且只有一条边连向它的父亲结点，诶？？有且只有一条？那！！我们就可以用这个点的点权去代替它与它父亲之间边的边权呀！！那这不就又成了树链剖分板子题了嘛？！\(QwQ\)。还有一个坑就是最后查询的时候，因为LCA的点权不在查询的路径范围内，因此要排除掉。
洛谷上此题的c++评测有些问题，下面给出c的代码(memset必须要全写，不然没法AC，我也不知道为什么)：
#include <ctype.h>
#include <stdio.h>
#include <limits.h>
#include <stdlib.h>
#include <string.h>
#define maxn 10007
#define ls rt<<1
#define rs rt<<1|1
int id[maxn],d[maxn],n,cnt,son[maxn],top[maxn],maxx[maxn<<2],t;
int head[maxn],num,fa[maxn],siz[maxn],w[maxn],a[maxn];
char s[8];
struct node {
int v,w,nxt;
}e[maxn<<1];
int max(int a, int b) { return a > b ? a : b; }
#define swap(A, B) \
{ \
int __T = A; \
A = B; \
B = __T; \
}
void ct(int u, int v, int w) {
e[++num].v=v;
e[num].w=w;
e[num].nxt=head[u];
head[u]=num;
}
void pushup(int rt) {
maxx[rt]=max(maxx[ls],maxx[rs]);
}
void build(int rt, int l, int r) {
if(l==r) {
maxx[rt]=a[l];
return;
}
int mid=(l+r)>>1;
build(ls,l,mid);
build(rs,mid+1,r);
pushup(rt);
}
void modify(int rt, int l, int r, int L, int val) {
if(l==r) {
maxx[rt]=val;
return;
}
int mid=(l+r)>>1;
if(L<=mid) modify(ls,l,mid,L,val);
else modify(rs,mid+1,r,L,val);
pushup(rt);
}
int cmax(int rt, int l, int r, int L, int R) {
if(L<=l&&r<=R) return maxx[rt];
int ans=0;
int mid=(l+r)>>1;
if(L<=mid) ans=max(ans,cmax(ls,l,mid,L,R));
if(R>mid) ans=max(ans,cmax(rs,mid+1,r,L,R));
return ans;
}
void dfs1(int u) {
siz[u]=1;
for(int i=head[u];i;i=e[i].nxt) {
int v=e[i].v;
if(v!=fa[u]) {
d[v]=d[u]+1;
fa[v]=u;
w[v]=e[i].w;
dfs1(v);
siz[u]+=siz[v];
if(siz[v]>siz[son[u]]) son[u]=v;
}
}
}
void dfs2(int u, int t) {
id[u]=++cnt;
top[u]=t;
a[cnt]=w[u];
if(son[u]) dfs2(son[u],t);
for(int i=head[u];i;i=e[i].nxt) {
int v=e[i].v;
if(v!=fa[u]&&v!=son[u]) dfs2(v,v);
}
}
int query(int x, int y) {
int ans=0,fx=top[x],fy=top[y];
while(fx!=fy) {
if(d[fx]<d[fy]) {
swap(x,y);
swap(fx,fy);
}
ans=max(ans,cmax(1,1,n,id[fx],id[x]));
x=fa[fx],fx=top[x];
}
if(id[x]>id[y]) swap(x,y);
ans=max(ans,cmax(1,1,n,id[x]+1,id[y]));
return ans;
}
int main() {
scanf("%d",&t);
while(t--) {
memset(head,0,sizeof(head));
memset(siz,0,sizeof(siz));
memset(id,0,sizeof(id));
memset(top,0,sizeof(top));
memset(son,0,sizeof(son));
memset(w,0,sizeof(w));
memset(maxx,0,sizeof(maxx));
memset(a,0,sizeof(a));
memset(fa,0,sizeof(fa));
memset(d,0,sizeof(d));num=0;cnt=0;
scanf("%d",&n);
for(int i=1,u,v,w;i<n;++i) {
scanf("%d%d%d",&u,&v,&w);
ct(u,v,w),ct(v,u,w);
}
dfs1(1);dfs2(1,1);build(1,1,n);
while(1) {
scanf("%s",s);
if(s[0]=='D') break;
if(s[0]=='C') {
int x,y;
scanf("%d%d",&x,&y);
x=d[e[x*2-1].v]<d[e[x<<1].v]?e[x<<1].v:e[x*2-1].v;
modify(1,1,cnt,id[x],y);
}
if(s[0]=='Q') {
int x,y;
scanf("%d%d",&x,&y);
printf("%d\n",query(x,y));
}
}
}
return 0;
}
接下来是在SPOJ上能够AC的c++代码：
#include<cstdio>
#include<algorithm>
#include<cstring>
#define maxn 10007
#define ls rt<<1
#define rs rt<<1|1
using namespace std;
int id[maxn],d[maxn],n,cnt,son[maxn],top[maxn],maxx[maxn<<2],t;
int head[maxn],num,fa[maxn],siz[maxn],w[maxn],a[maxn];
char s[8];
struct node {
int v,w,nxt;
}e[maxn<<1];
inline void ct(int u, int v, int w) {
e[++num].v=v;
e[num].w=w;
e[num].nxt=head[u];
head[u]=num;
}
inline void pushup(int rt) {
maxx[rt]=max(maxx[ls],maxx[rs]);
}
void build(int rt, int l, int r) {
if(l==r) {
maxx[rt]=a[l];
return;
}
int mid=(l+r)>>1;
build(ls,l,mid);
build(rs,mid+1,r);
pushup(rt);
}
void modify(int rt, int l, int r, int L, int val) {
if(l==r) {
maxx[rt]=val;
return;
}
int mid=(l+r)>>1;
if(L<=mid) modify(ls,l,mid,L,val);
else modify(rs,mid+1,r,L,val);
pushup(rt);
}
int cmax(int rt, int l, int r, int L, int R) {
if(L<=l&&r<=R) return maxx[rt];
int ans=0;
int mid=(l+r)>>1;
if(L<=mid) ans=max(ans,cmax(ls,l,mid,L,R));
if(R>mid) ans=max(ans,cmax(rs,mid+1,r,L,R));
return ans;
}
void dfs1(int u) {
siz[u]=1;
for(int i=head[u];i;i=e[i].nxt) {
int v=e[i].v;
if(v!=fa[u]) {
d[v]=d[u]+1;
fa[v]=u;
w[v]=e[i].w;
dfs1(v);
siz[u]+=siz[v];
if(siz[v]>siz[son[u]]) son[u]=v;
}
}
}
void dfs2(int u, int t) {
id[u]=++cnt;
top[u]=t;
a[cnt]=w[u];
if(son[u]) dfs2(son[u],t);
for(int i=head[u];i;i=e[i].nxt) {
int v=e[i].v;
if(v!=fa[u]&&v!=son[u]) dfs2(v,v);
}
}
int query(int x, int y) {
int ans=0,fx=top[x],fy=top[y];
while(fx!=fy) {
if(d[fx]<d[fy]) {
swap(x,y);
swap(fx,fy);
}
ans=max(ans,cmax(1,1,n,id[fx],id[x]));
x=fa[fx],fx=top[x];
}
if(id[x]>id[y]) swap(x,y);
ans=max(ans,cmax(1,1,n,id[x]+1,id[y]));
return ans;
}
int main() {
scanf("%d",&t);
while(t--) {
memset(head,0,sizeof(head));
memset(siz,0,sizeof(siz));
memset(id,0,sizeof(id));
memset(top,0,sizeof(top));
memset(son,0,sizeof(son));
memset(w,0,sizeof(w));
memset(maxx,0,sizeof(maxx));
memset(a,0,sizeof(a));
memset(fa,0,sizeof(fa));
memset(d,0,sizeof(d));num=0;cnt=0;
scanf("%d",&n);
for(int i=1,u,v,w;i<n;++i) {
scanf("%d%d%d",&u,&v,&w);
ct(u,v,w),ct(v,u,w);
}
dfs1(1);dfs2(1,1);build(1,1,n);
while(1) {
scanf("%s",s);
if(s[0]=='D') break;
if(s[0]=='C') {
int x,y;
scanf("%d%d",&x,&y);
x=d[e[x*2-1].v]<d[e[x<<1].v]?e[x<<1].v:e[x*2-1].v;
modify(1,1,cnt,id[x],y);
}
if(s[0]=='Q') {
int x,y;
scanf("%d%d",&x,&y);
printf("%d\n",query(x,y));
}
}
}
return 0;
}
孤注一掷，我怎甘落空？！
