| Time Limit: 4000MS | Memory Limit: 65536K | |
| Total Submissions: 7918 | Accepted: 2078 |
Description
After their royal wedding, Jiajia and Wind hid away in XX Village, to enjoy their ordinary happy life. People in XX Village lived in beautiful huts. There are some pairs of huts connected by bidirectional roads. We say that huts in the same pair directly connected.
XX Village is so special that we can reach any other huts starting from an arbitrary hut. If each road cannot be walked along twice, then the route between every pair is unique.
Since Jiajia earned enough money, Wind became a housewife. Their children loved to go to other kids, then make a simple call to Wind: 'Mummy, take me home!'
At different times, the time needed to walk along a road may be different. For example, Wind takes 5 minutes on a road normally, but may take 10 minutes if there is a lovely little dog to play with, or take 3 minutes if there is some unknown strange smell surrounding the road.
Wind loves her children, so she would like to tell her children the exact time she will spend on the roads. Can you help her?
Since Jiajia earned enough money, Wind became a housewife. Their children loved to go to other kids, then make a simple call to Wind: 'Mummy, take me home!'
At different times, the time needed to walk along a road may be different. For example, Wind takes 5 minutes on a road normally, but may take 10 minutes if there is a lovely little dog to play with, or take 3 minutes if there is some unknown strange smell surrounding the road.
Wind loves her children, so she would like to tell her children the exact time she will spend on the roads. Can you help her?
Input
The first line contains three integers n, q, s. There are n huts in XX Village, q messages to process, and Wind is currently in hut s. n < 100001 , q < 100001.
The following n-1 lines each contains three integers a, b and w. That means there is a road directly connecting hut a and b, time required is w. 1<=w<= 10000.
The following q lines each is one of the following two types:
Message A: 0 u
A kid in hut u calls Wind. She should go to hut u from her current position.
Message B: 1 i w
The time required for i-th road is changed to w. Note that the time change will not happen when Wind is on her way. The changed can only happen when Wind is staying somewhere, waiting to take the next kid.
The following n-1 lines each contains three integers a, b and w. That means there is a road directly connecting hut a and b, time required is w. 1<=w<= 10000.
The following q lines each is one of the following two types:
Message A: 0 u
A kid in hut u calls Wind. She should go to hut u from her current position.
Message B: 1 i w
The time required for i-th road is changed to w. Note that the time change will not happen when Wind is on her way. The changed can only happen when Wind is staying somewhere, waiting to take the next kid.
Output
For each message A, print an integer X, the time required to take the next child.
Sample Input
3 3 1 1 2 1 2 3 2 0 2 1 2 3 0 3
Sample Output
1
3
这题可以用裸的在线lca做，也可以用树状数组做。
代码一：裸的在线lca，每次更改边的权值，就把该点的子树重新更新一遍。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x7fffffff
#define maxn 100010
struct node{
int to,len,next,from;
}e[2*maxn];
int first[maxn],dep[maxn],f[maxn][30],dist[maxn],pre[maxn];
int lca(int x,int y){
int i;
if(dep[x]<dep[y]){
swap(x,y);
}
for(i=18;i>=0;i--){
if(dep[f[x][i] ]>=dep[y]){
x=f[x][i];
}
}
if(x==y)return x;
for(i=18;i>=0;i--){
if(f[x][i]!=f[y][i]){
x=f[x][i];y=f[y][i];
}
}
return f[x][0];
}
void dfs(int x,int fa){
int i,y;
dep[x]=dep[fa]+1;
for(i=first[x];i!=-1;i=e[i].next){
y=e[i].to;
if(y!=fa){
f[y][0]=x;
pre[y]=x;
dist[y]=dist[x]+e[i].len;
dfs(y,x);
}
}
}
int main()
{
int n,m,i,j,T,q,s,d,c,g,tot,k;
while(scanf("%d%d%d",&n,&q,&s)!=EOF)
{
tot=0;
memset(first,-1,sizeof(first));
for(i=1;i<=n-1;i++){
scanf("%d%d%d",&c,&d,&g);
e[i].next=first[c];e[i].len=g;e[i].to=d;e[i].from=c;
first[c]=i;
e[i+n-1].next=first[d];e[i+n-1].len=g;e[i+n-1].to=c;e[i+n-1].from=d;
first[d]=i+n-1;
}
pre[1]=0;
dep[0]=0;dist[1]=0;
dfs(1,0);
for(k=1;k<=18;k++){
for(i=1;i<=n;i++){
f[i][k]=f[f[i][k-1]][k-1];
}
}
for(i=1;i<=q;i++){
scanf("%d%d",&g,&c);
if(g==0){
int t=lca(s,c);
printf("%d
",dist[s]+dist[c]-2*dist[t]);
s=c;
}
else{
scanf("%d",&d);
int u=e[c].from;
int v=e[c].to;
int t1,t2;
e[c].len=e[c+n-1].len=d;
if(pre[u]==v){
t1=u;t2=v;
}
else{
t1=v;t2=u;
}
dist[t1]=dist[t2]+d;
dfs(t1,t2);
}
}
}
return 0;
}代码二：树上两个节点a,b的距离可以转化为dis[a] + dis[b] - 2*dis[lca(a,b)]，其中 dis[i] 表示 i 节点到根的距离，由于每次修改一条边，树中在这条边下方的 dis[] 值全都会受到影响，这样每条边都对应这一段这条边的管辖区，可以深搜保存遍历该点的时间戳，st[i] 表示第一次遍历到该点的时间戳,ed[i] 表示回溯到该点时的时间戳，这样每次修改边 i 的时候就可以对区间[st[i],ed[i]]进行成段更新，即位置 ll[i] 上加一个权值，在位置rr[i]+1 上减去这个权值，求和时，sum(ll[i]) 即为该点到根的距离。这里注意初始化的时候，可以把输入的每一条边都当做更新的边，这样初始操作和更新操作都是一样的。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define inf 0x7fffffff
#define maxn 100050
struct node{
int len,to,next,from;
}e[2*maxn];
int first[maxn],dep[maxn],f[maxn][30],b[maxn],st[maxn],ed[maxn],tot,pre[maxn];
int lowbit(int x){
return x&(-x);
}
void update(int pos,int num){
while(pos<maxn){
b[pos]+=num;pos+=lowbit(pos);
}
}
int getsum(int pos){
int num=0;
while(pos>0){
num+=b[pos];pos-=lowbit(pos);
}
return num;
}
void dfs(int x,int fa){
int i,y;
dep[x]=dep[fa]+1;
st[x]=tot;
for(i=first[x];i!=-1;i=e[i].next){
y=e[i].to;
if(y!=fa){
f[y][0]=x;
pre[y]=x;
tot++;
dfs(y,x);
}
}
ed[x]=tot;
}
int lca(int x,int y){
int i;
if(dep[x]<dep[y]){
swap(x,y);
}
for(i=18;i>=0;i--){
if(dep[f[x][i] ]>=dep[y]){
x=f[x][i];
}
}
if(x==y)return x;
for(i=18;i>=0;i--){
if(f[x][i]!=f[y][i]){
x=f[x][i];y=f[y][i];
}
}
return f[x][0];
}
int main()
{
int n,m,i,j,q,s,c,d,g,k,u,v;
while(scanf("%d%d%d",&n,&q,&s)!=EOF)
{
memset(first,-1,sizeof(first));
for(i=1;i<=n-1;i++){
scanf("%d%d%d",&c,&d,&g);
e[i].len=g;e[i].next=first[c];e[i].to=d;e[i].from=c;
first[c]=i;
e[i+n-1].len=g;e[i+n-1].next=first[d];e[i+n-1].to=c;e[i+n-1].from=d;
first[d]=i+n-1;
}
memset(b,0,sizeof(b));
dep[0]=0;
tot=1;
dfs(1,0);
for(k=1;k<=18;k++){
for(i=1;i<=n;i++){
f[i][k]=f[f[i][k-1]][k-1];
}
}
for(c=1;c<=n-1;c++){
u=e[c].from;v=e[c].to;
if(pre[v]==u){
swap(u,v);
}
update(st[u],e[c].len); //这里只要更新深度更深的点的子树就行。
update(ed[u]+1,-e[c].len);
}
for(i=1;i<=q;i++){
scanf("%d",&g);
if(g==0){
scanf("%d",&c);
int t=lca(s,c);
printf("%d
",getsum(st[s])+getsum(st[c])-2*getsum(st[t]) );
s=c;
}
else{
scanf("%d%d",&c,&d);
u=e[c].from;v=e[c].to;
if(pre[v]==u){
swap(u,v);
}
int cha=d-e[c].len;
e[c].len=e[c+n-1].len=d;
update(st[u],cha);
update(ed[u]+1,-cha);
}
}
}
return 0;
}