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
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1 #include<cstdio> 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 ",sum(tid[k])+a[k]); 55 } 56 } 57 return 0; 58 }