POJ 2763 Housewife Wind

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?

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【题目分析】
树链剖分+线段树

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【代码】

#include <cstdio>#include <cmath>#include <cstring>#include <iostream>#include <algorithm>using namespace std;int n,cur,q,root=1,node=0,opt,x,y;int dep[100001],son[100001],size[100001],fa[100001],pos[100001],v[100001],top[100001];int data[100001];int h[200001],to[200001],ne[200001],w[200001],en=0;struct node{    int l,r,sum;}t[100001*8];void add(int a,int b,int c){    ne[en]=h[a];    to[en]=b;    w[en]=c;    h[a]=en++;}void dfs(int k){    size[k]=1;    for (int i=h[k];i>=0;i=ne[i])    {        if (to[i]!=fa[k])        {            fa[to[i]]=k;            dep[to[i]]=dep[k]+1;            v[to[i]]=w[i];            dfs(to[i]);            size[k]+=size[to[i]];            if (size[to[i]]>size[son[k]]) son[k]=to[i];        }    }}void dfs(int k,int tp){    top[k]=tp;    pos[k]=++node;    data[node]=v[k];    if (son[k]) dfs(son[k],tp);    for (int i=h[k];i>=0;i=ne[i])        if (to[i]!=fa[k]&&to[i]!=son[k])            dfs(to[i],to[i]);}void build(int k,int L,int R){    t[k].l=L;    t[k].r=R;    if (L==R)    {        t[k].sum=data[L];        return ;    }    build(k*2,L,(L+R)/2);    build(k*2+1,(L+R)/2+1,R);    t[k].sum=t[k*2].sum+t[k*2+1].sum;}int qsum(int k,int L,int R){    if (t[k].l>R||t[k].r<L) return 0;    if (t[k].l>=L&&t[k].r<=R) return t[k].sum;    return qsum(k*2,L,R)+qsum(k*2+1,L,R);}void mod(int k,int x){    if (t[k].l==t[k].r)    {        t[k].sum=data[t[k].l];        return ;    }    if (x<=(t[k].l+t[k].r)/2) mod(k*2,x);    else mod(k*2+1,x);    t[k].sum=t[k*2].sum+t[k*2+1].sum;}int main(){    memset(h,-1,sizeof h);    scanf("%d%d%d",&n,&q,&cur);    for (int i=1;i<n;++i)    {        int a,b,c;        scanf("%d%d%d",&a,&b,&c);        add(a,b,c);        add(b,a,c);    }    dfs(root);    dfs(root,root);    build(1,1,node);    while (q--)    {        scanf("%d",&opt);        if (opt==0)        {            int ans=0;            scanf("%d",&x);            int a=cur,b=x; cur=x;            while (top[a]!=top[b])            {                if (dep[top[a]]<dep[top[b]]) swap(a,b);                ans+=qsum(1,pos[top[a]],pos[a]);                a=fa[top[a]];            }            if (dep[a]<dep[b]) swap(a,b);            if (a!=b) ans+=qsum(1,pos[b]+1,pos[a]);            printf("%d\n",ans);        }        else        {            scanf("%d%d",&x,&y);            int a=to[(x-1)*2],b=to[(x-1)*2+1];//          printf("%d %d\n",a,b);            if (dep[a]<dep[b]) swap(a,b);            data[pos[a]]=y;            mod(1,pos[a]);        }    }}