题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=3966

题意：给你一棵树，然后给定点之间的路径权值修改，最后单点查询

题解：树链剖分裸题，这里我用树状数组维护

1 #include
      
        2 #pragma comment(linker, "/STACK:1024000000,1024000000") 3 #define F(i,a,b) for(int i=a;i<=b;++i) 4  5 const int N=50010;char op[2]; 6 int n,m,p,tp,x,y,k,a[N],tree[N],nxt[2*N],g[N],v[2*N],ed,hs[N],fa[N],top[N],dep[N],siz[N],idx,tid[N]; 7  8 inline void adg(int x,int y){v[++ed]=y,nxt[ed]=g[x],g[x]=ed;} 9 //树状树组部分10 inline void update(int x,int k){
    while(x<=n)tree[x]+=k,x+=x&-x;}11 inline int sum(int x){
    int ans=0;for(;x>0;x-=x&-x)ans+=tree[x];return ans;}12 //树链部分13 void dfs1(int u,int pre){14     fa[u]=pre,siz[u]=1,dep[u]=dep[pre]+1,hs[u]=0;15     for(int i=g[u];~i;i=nxt[i]){16         int vv=v[i];17         if(vv!=pre){18             dfs1(vv,u);19             if(siz[vv]>siz[hs[u]])hs[u]=vv;20             siz[u]+=siz[vv];21         }22     }23 }24 25 void dfs2(int u,int tp){26     tid[u]=++idx,top[u]=tp;27     if(hs[u])dfs2(hs[u],tp);28     for(int i=g[u];~i;i=nxt[i]){29         int vv=v[i];30         if(vv!=fa[u]&&vv!=hs[u])dfs2(vv,vv);31     }32 }33 34 void up(int x,int y,int k){35     int fx=top[x],fy=top[y];36     while(fx!=fy){37         if(dep[fx]>=dep[fy])update(tid[fx],k),update(tid[x]+1,-k),x=fa[fx],fx=top[x];38         else update(tid[fy],k),update(tid[y]+1,-k),y=fa[fy],fy=top[y];39     }40     if(dep[x]>dep[y])x=x^y,y=x^y,x=x^y;41     update(tid[x],k),update(tid[y]+1,-k);42 }43 44 int main(){45     while(~scanf("%d%d%d",&n,&m,&p)){46         F(i,1,n)scanf("%d",a+i);47         F(i,0,N-1)g[i]=-1,tree[i]=0;ed=0;48         F(i,1,m)scanf("%d%d",&x,&y),adg(x,y),adg(y,x);49         dfs1(1,0),idx=0,dfs2(1,1);50         F(i,1,p){51             scanf("%s",op);52             if(op[0]=='I')scanf("%d%d%d",&x,&y,&k),up(x,y,k);53             else if(op[0]=='D')scanf("%d%d%d",&x,&y,&k),up(x,y,-k);54             else scanf("%d",&k),printf("%d\n",sum(tid[k])+a[k]);55         }56     }57     return 0;58 }