POJ2763-Housewife Wind-树上单点修改区间求和

这道题可以树链剖分做。但是最近在给学弟搞数据结构复习了LCA树状数组RMQ

然后就搞了一发LCA+树状数组维护。

dis数组维护当前点到根节点的权值和。则dis(u,v) = dis[u]+dis[v]-2*dis[lca(u,v)]

修改的时候，单点修改影响了该点所有儿子的dis，刚好可以用dfs序定位所有儿子，用树状数组改段求点很方便。

http://blog.csdn.net/q573290534/article/details/6664454 树状数组的三种求和（改点求段，改段求点，改段求段）

http://blog.csdn.net/qwe2434127/article/details/49819975 dfs序题目类型

http://www.cnblogs.com/longdouhzt/archive/2011/10/16/2214034.html LCA题目小节

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

using namespace std;

const int maxn = 2e5+100;

int N,Q,S;
//vector <int> G[maxn];

struct edge{
    int to,next,w;
}e[2*maxn];

int head[2*maxn],tot;
void add_edge(int u,int v,int w)
{
    e[tot].to = v;
    e[tot].w = w;
    e[tot].next = head[u];
    head[u] = tot++;

    e[tot].to = u;
    e[tot].w = w;
    e[tot].next = head[v];
    head[v] = tot++;
}

int in[maxn],out[maxn],P[2*maxn],fa[maxn][30],dep[maxn],dis[maxn],cnt;
void dfs(int u,int _fa,int _dep,int _dis)
{
    in[u] = ++cnt;
    P[cnt] = u;
    fa[u][0] = _fa;
    dis[u] = _dis;
    dep[u] = _dep;
    for(int i=head[u];~i;i=e[i].next)
    {
        int v = e[i].to;
        if(v == _fa) continue;
        dfs(v,u,_dep+1,_dis+e[i].w);
    }
    out[u] = ++cnt;
}
void debug()
{
    printf("in:\t");for(int i=1;i<=N;i++) printf("%d ",in[i]);puts("");
    printf("out:\t");for(int i=1;i<=N;i++) printf("%d ",out[i]);puts("");
    printf("p:\t");for(int i=1;i<=cnt;i++) printf("%d ",P[i]);puts("");
    printf("dis:\t");for(int i=1;i<=N;i++) printf("%d ",dis[i]);puts("");
    printf("dep:\t");for(int i=1;i<=N;i++) printf("%d ",dep[i]);puts("");
    printf("edge:");for(int i=0;i<tot;i++) printf("\tto:%d w:%d\n",e[i].to,e[i].w);
}

int initLCA()
{
    int m = (int)log(N)/log(2)+1;
    for(int k=0;k<m;k++)
    {
        for(int v=1;v<=N;v++)
        {
            if(fa[v][k] < 0) {fa[v][k+1] = -1;continue;}
            else fa[v][k+1] = fa[fa[v][k]][k];
        }
    }
}

int LCA(int u,int v)
{
    int m = (int)log(N)/log(2)+1;
    if(dep[v] > dep[u]) swap(u,v);
    for(int k=0;k<m;k++)
    {
        if((dep[u]-dep[v])>>k & 1 )
            u = fa[u][k];
    }
    if(u == v) return u;
    for(int k=m-1;k>=0;k--)
    {
        if(fa[u][k] != fa[v][k])
        {
            u = fa[u][k];
            v = fa[v][k];
        }
    }
    return fa[u][0];
}

int c[2*maxn];
int lowbit(int x) {return x&-x;}

void init()
{
    memset(head,-1,sizeof head);
    memset(fa,-1,sizeof fa);
    memset(c,0,sizeof c);
    memset(in,0,sizeof in);
    memset(out,0,sizeof out);
    tot = 0;
    cnt = 0;
}

void add(int x,int d)
{
    while(x)
    {
        c[x] += d;
        x -= lowbit(x);
    }
}
void add_seg(int l,int r,int d)
{
    add(r,d);
    add(l-1,-d);
}
int sum(int x)
{
    int res = 0;
    //printf("u:%d ",P[x]);
    while(x <= cnt)
    {
        res += c[x];
        x += lowbit(x);
    }
    //printf("res:%d\n",res);
    return res;
}
int dist(int x)
{
    if(x == -1) return 0;
    return sum(in[x]) + dis[x];
}
int main()
{
    //freopen("input.txt","r",stdin);
    while(~scanf("%d%d%d",&N,&Q,&S))
    {
        init();
        for(int i=0,u,v,w;i<N-1;i++)
        {
            scanf("%d%d%d",&u,&v,&w);
            add_edge(u,v,w);
        }
        dfs(1,-1,1,0);
        //debug();
        initLCA();
        for(int i=0;i<tot;i+=2)
        {
            if(dep[e[i].to] < dep[e[i+1].to]) swap(e[i].to,e[i+1].to);
        }
        int op;
        for(int i=0,a,b,c;i<Q;i++)
        {
            scanf("%d",&op);
            if(op == 0)
            {
                scanf("%d",&a);
                int lca = LCA(S,a);
                //printf("lca:%d\n",lca);

                printf("%d\n",dist(S)+dist(a)-2*(dist(lca)));
                S = a;
            }else
            {
                scanf("%d%d",&a,&b);
                a--;
                int u = e[a*2].to;
                int dw = b - (dist(u) - dist(fa[u][0]));
                //printf("dw:%d u:%d\n",dw,u);
                add_seg(in[u],out[u],dw);
            }
        }
    }
}