POJ 2763 Housewife Wind 树链剖分

来源：互联网发布：ios java jit 编辑：程序博客网时间：2024/05/21 06:18

【题目大意】

有一棵树，每条边有边权。你最开始位于点s。有q次操作。0 u：你从s走到u，输出这条简单路径的边权和，并令 s = u。1 i w，把输入的第i条边的边权变为w。

【思路】

没有多想，直接用剖分水过。。。

//#pragma comment(linker, "/STACK:102400000,102400000")#include<cstdio>#include<cstring>#include<vector>#include<queue>#include<cmath>#include<cctype>#include<string>#include<algorithm>#include<iostream>#include<ctime>#include<map>#include<set>using namespace std;#define MP(x,y) make_pair((x),(y))#define PB(x) push_back(x)//typedef __int64 LL;//typedef unsigned __int64 ULL;/* ****************** */const int INF=1000111222;const double INFF=1e200;const double eps=1e-8;const int mod=1000000007;const int NN=100010;const int MM=100010;/* ****************** */struct G{    int v,next,w,id;}E[NN*2];int p[NN],T;int si[NN],per[NN],deep[NN],anc[NN];int tsp,pos[NN],top[NN];int val[NN],a[NN];struct TR{    int l,r,sum;    int mid()    {        return (l+r)>>1;    }}tr[NN*4];void add(int u,int v,int w){    E[T].v=v;    E[T].w=w;    E[T].next=p[u];    p[u]=T++;}void c_dfs1(int u,int fa,int cen){    anc[u] = fa;    deep[u] = cen;    si[u] = 1;    per[u] = -1;    int i,v;    for(i=p[u];i+1;i=E[i].next)    {        v=E[i].v;        if(v==fa)continue;        val[v] = E[i].w;        E[i].id = E[i^1].id= v;        c_dfs1(v,u,cen+1);        si[u]+=si[v];        if(per[u]==-1 || si[v]>si[ per[u] ])            per[u]=v;    }}void c_dfs2(int u,int fa,int now_top){    pos[u]  = ++tsp;    top[u] = now_top;    a[tsp] = val[u];    if(per[u]!=-1)c_dfs2(per[u],u,now_top);    int i,v;    for(i=p[u];i+1;i=E[i].next)    {        v = E[i].v;        if(v==fa || v==per[u])continue;        c_dfs2(v,u,v);    }}void push_up(int R){    tr[R].sum = tr[R<<1].sum + tr[R<<1|1].sum;}void build(int l,int r,int R){    tr[R].l=l;    tr[R].r=r;    if(l==r)    {        tr[R].sum = a[l];        return;    }    int mid=tr[R].mid();    build(l,mid,R<<1);    build(mid+1,r,R<<1|1);    push_up(R);}void update(int x,int R,int col){    if(tr[R].l==tr[R].r)    {        tr[R].sum = col;        return;    }    int mid=tr[R].mid();    if(x<=mid)        update(x,R<<1,col);    else        update(x,R<<1|1,col);    push_up(R);}int query(int l,int r,int R){    if(l<=tr[R].l && tr[R].r<=r)        return tr[R].sum;    int mid=tr[R].mid();    int sum=0;    if(l<=mid)        sum += query(l,r,R<<1);    if(r>=mid+1)        sum += query(l,r,R<<1|1);    return sum;}void solve(int u,int v){    int ans = 0;    while(top[u] != top[v])    {        if( deep[ top[u] ] > deep[ top[v] ] )        {            ans += query(pos[ top[u] ],pos[u],1);            u = anc[ top[u] ];        }        else        {            ans += query(pos[ top[v] ],pos[v],1);            v = anc[ top[v] ];        }    }    if(pos[u]>pos[v])        swap(u,v);    ans +=  query(pos[u],pos[v],1);    ans -= val[u];    printf("%d\n",ans);}int main(){    int n,m,s,u,v,w,i,op,id;    while(scanf("%d%d%d",&n,&m,&s)!=EOF)    {        memset(p,-1,sizeof(p));        T=0;        for(i=1;i<n;i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v,w);            add(v,u,w);        }        val[1] = 0;        c_dfs1(1,-1,0);        tsp = 0;        c_dfs2(1,-1,1);        build(1,n,1);        u = s;        for(i=0;i<m;i++)        {            scanf("%d",&op);            if(op==0)            {                scanf("%d",&v);                solve(u,v);                u = v;            }            else            {                scanf("%d%d",&id,&w);                update( pos[ E[(id-1)*2].id ] , 1, w );                val[ E[(id-1)*2].id ] = w;            }        }    }    return 0;}

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