POJ 2763 Housewife Wind LCA基础题

Time Limit: 4000MS		Memory Limit: 65536K
Total Submissions: 7047		Accepted: 1784

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? 

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. 

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

题意：一个无根树，给出主角一开始所在的位置S，然后下面q个操作，操作包括查询和修改操作，对于查询操作就是当前主角的位置到目的点的距离是多少，然后主角去到那里之后就在那里等待，下次查询的时候那里就是新的起点（所以sample中第二次查询为什么是3）。修改是修改第k条边的权值，边的编号就是输入的顺序。

这题可能是数据水了还是怎么，对于修改操作虽然有优化的方法，但是用最朴素的直接遍历下去修改也是可行的，不会超时，不过时间就比较糟糕了

修改操作其实是修改了一部分子树的sum值，对于查询操作就是普通的LCA

注意：
LCA + 修改边权：一边查询两点间的距离，一边修改某些边权。对于修改了某些边的边权，就要从此开始遍历下面的子孙后代更改他们的sum值（点到根的距离）。

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

const int maxn=100000+5;

struct Edge
{
    int from,to,next,w;
}edge[maxn<<1];

int head[maxn];
int tot;

int dep[maxn];
int p[maxn][25];
int sum[maxn];

void init()
{
    memset(head,-1,sizeof(head));
    memset(dep,0,sizeof(dep));
    memset(p,-1,sizeof(p));
    memset(sum,0,sizeof(sum));
    tot=1;
}

void addedge(int u,int v,int w)
{
    edge[tot].from=u;
    edge[tot].to=v;
    edge[tot].w=w;
    edge[tot].next=head[u];
    head[u]=tot++;
}

void dfs(int u,int fa)
{
    for(int i=head[u];~i;i=edge[i].next)
    {
        int v=edge[i].to;
        int w=edge[i].w;
        if(!dep[v]&&v!=fa)
        {
            dep[v]=dep[u]+1;
            p[v][0]=u;
            sum[v]=sum[u]+w;
            dfs(v,u);
        }
    }
}

void dfs2(int u,int fa)
{
    for(int i=head[u];~i;i=edge[i].next)
    {
        int w=edge[i].w;
        int v=edge[i].to;
        if(v==fa)
            continue;
        sum[v]=sum[u]+w;
        dfs2(v,u);
    }
}

void init_lca(int n)
{
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=1;i<=n;i++)
        {
            if(p[i][j-1]!=-1)
                p[i][j]=p[p[i][j-1]][j-1];
        }
    }
}

void update(int u,int v,int b)
{
    if(dep[u]>dep[v])
        swap(u,v);
    sum[v]=sum[u]+b;
    dfs2(v,u);
}

int solve(int n,int u,int v)
{
    if(dep[u]<dep[v])
        swap(u,v);

    int init_u=u;
    int init_v=v;

    int cnt;
    for(cnt=0;(1<<cnt)<=dep[u];cnt++)
        ;
    cnt--;

    for(int j=cnt;j>=0;j--)
    {
        if(dep[u]-(1<<j)>=dep[v])
            u=p[u][j];
    }

    if(u==v)
        return (sum[init_u]-sum[v]);

    int lca;

    for(int j=cnt;j>=0;j--)
    {
        if(p[u][j]!=-1&&p[u][j]!=p[v][j])
        {
            u=p[u][j];
            v=p[v][j];
        }
    }

    lca=p[u][0];

    return (sum[init_u]+sum[init_v]-2*sum[lca]);
}

int main()
{
    int n,q,s;
    while(scanf("%d%d%d",&n,&q,&s)!=EOF)
    {
        init();

        for(int i=1;i<n;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }

        dfs(1,-1);

        init_lca(n);

        /*
        for(int i=1;i<=n;i++)
            printf("%d\n",sum[i]);
        */

        for(int i=0;i<q;i++)
        {
            int k;
            scanf("%d",&k);
            if(k==1)
            {
                int a,b;
                scanf("%d%d",&a,&b);
                edge[2*a-1].w=b;
                edge[2*a].w=b;
                update(edge[2*a].from,edge[2*a].to,b);
                
                /*
                for(int i=1;i<=n;i++)
                    printf("%d\n",sum[i]);
                */
                
            }
            else
            {
                int a;
                scanf("%d",&a);
                printf("%d\n",solve(n,s,a));
                s=a;
            }
        }
    }

    return 0;
}