poj 2763 Housewife Wind(树链剖分)

题意：给出一个n个节点的树，每次询问树上路径长度，或者修改某条边的长度。

思路：典型的树链剖分，但是我又写了好久。。。唉，还是代码能力不够啊，每次写都要写出好多bug……

代码：

#include<iostream>#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<map>#include<queue>#include<set>#include<stack>#include<cmath>#include<vector>#define inf 0x3f3f3f3f#define Inf 0x3FFFFFFFFFFFFFFFLL#define eps 1e-9#define pi acos(-1.0)using namespace std;typedef long long ll;const int maxn=100000+10;struct Edge{    int u,v,w,next;    Edge (){};    Edge (int uu,int vv,int ww,int nx) {u=uu;v=vv;w=ww;next=nx;}}edges[maxn<<1];int head[maxn],d[maxn],nEdge,N;int childs[maxn],pa[maxn],ft[maxn],id[maxn<<1],val[maxn];bool flag[maxn<<1];void AddEdges(int u,int v,int w){    edges[++nEdge]=Edge(u,v,w,head[u]);    head[u]=nEdge;    edges[++nEdge]=Edge(v,u,w,head[v]);    head[v]=nEdge;}void Init(){    memset(head,0xff,sizeof(head));    memset(d,0,sizeof(d));    memset(flag,0,sizeof(flag));    memset(ft,0xff,sizeof(ft));    nEdge=-1;N=0;}//划分轻重边int dfs(int u,int fa){    childs[u]=0;    int cc=1;    int maxb=-1,pos;    for(int k=head[u];k!=-1;k=edges[k].next)    {        int v=edges[k].v;        if(v==fa) continue;        pa[v]=k;childs[u]++;        d[v]=d[u]+1;        int cv=dfs(v,u);        cc+=cv;        if(cv>maxb) {maxb=cv;pos=k;}    }    if(maxb!=-1) flag[pos]=flag[pos^1]=true;    return cc;}void dfs2(int u,int fa){    if(fa==-1) ft[u]=-1;    else    {        if(flag[fa])        {            id[fa]=id[fa^1]=++N;val[N]=edges[fa].w;            if(childs[edges[fa].u]<=1&&ft[edges[fa].u]!=-1) ft[u]=ft[edges[fa].u];            else ft[u]=fa;        }        else ft[u]=-1;    }    for(int k=head[u];k!=-1;k=edges[k].next)    {        if(k==(fa^1)) continue;        dfs2(edges[k].v,k);    }}//线段树部分int sum[maxn<<2];inline void PushUp(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void build(int l,int r,int rt){    if(l==r)    {        sum[rt]=val[l];        return ;    }    int m=(l+r)>>1;    build(l,m,rt<<1);    build(m+1,r,rt<<1|1);    PushUp(rt);}int Query(int L,int R,int l,int r,int rt){    if(l>=L&&r<=R)    {        return sum[rt];    }    int m=(l+r)>>1;    int res=0;    if(m>=L) res+=Query(L,R,l,m,rt<<1);    if(m<R) res+=Query(L,R,m+1,r,rt<<1|1);    return res;}void Update(int p,int l,int r,int rt,int v){    if(l==r)    {        sum[rt]=v;        return ;    }    int m=(l+r)>>1;    if(m>=p) Update(p,l,m,rt<<1,v);    else Update(p,m+1,r,rt<<1|1,v);    PushUp(rt);}int solve(int x,int y){    if(x==y) return 0;    int res=0,L,R;    while(x!=y)    {        if(d[x]<d[y]) swap(x,y);        if(ft[x]!=-1)        {            L=id[ft[x]];R=id[pa[x]];            if(ft[y]==ft[x])            {                L=id[pa[y]]+1;                x=y;            }            else x=edges[ft[x]].u;            res+=Query(L,R,1,N,1);        }        else        {            res+=edges[pa[x]].w;            x=edges[pa[x]].u;        }    }    return res;}void change(int v,int w){    if(flag[v*2]) Update(id[v*2],1,N,1,w);    else edges[v*2].w=edges[(v*2)^1].w=w;}int main(){    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int n,q,s;    scanf("%d%d%d",&n,&q,&s);    Init();    int u,v,w;    for(int i=1;i<n;++i)    {        scanf("%d%d%d",&u,&v,&w);        AddEdges(u,v,w);    }    dfs(1,-1);dfs2(1,-1);    if(N) build(1,N,1);    int type,ans;    while(q--)    {        scanf("%d",&type);        if(type==0)        {            scanf("%d",&v);            ans=solve(s,v);            printf("%d\n",ans);            s=v;        }        else        {            scanf("%d%d",&v,&w);            change(v-1,w);        }    }    return 0;}