"BZOJ4034" [HAOI2015] tree operation tree chain split + segment tree

Action on "BZOJ4034" [HAOI2015] Tree description

There is a tree with a number of N, the root of point 1, and the tree point has edge rights. And then there's M-a

Operation, divided into three kinds: Operation 1: The point of a node x to increase the right of a. Action 2: Add a to the point in a subtree that has a node x as its root. Action 3: Ask a node x to the root of the path in a little bit of the right and. Input

The first line consists of two integers N, M. Represents the number of points and operands.

The next row of N integers represents the initial weights of the nodes in the tree. Next N-1 line three positive integer fr, to, indicating that there is an edge in the tree (FR, to). The next M line, each line represents one operation. The first number represents the type of operation (1-3) followed by the parameters of the operation (x or x a). Output

For each query operation, output the answer to the query. The answers are separated by line breaks.

Sample Input5 5
1 2 3) 4 5
1 2
1 4
2 3
2 5
3 3
1 2 1
3 5
2 1 2
3 3Sample Output6
9
13
HINT

For 100% of data, n,m<=100000, and the absolute value of all input data is not

Will exceed 10^6. Solving: Tree section template problem, note to open long long

#include <stdio.h>#include<string.h>#include<iostream>#defineLson x<<1#defineRson x<<1|1using namespaceStd;typedefLong Longll;Const intmaxn=100010;intn,m,cnt,tot;ll S[MAXN<<2],tag[maxn<<2];intto[maxn<<1],next[maxn<<1],V[MAXN],U[MAXN],FA[MAXN],HEAD[MAXN];intDEEP[MAXN],SIZE[MAXN],SON[MAXN],P[MAXN],TOP[MAXN];Charstr[Ten];intReadin () {intret=0, sig=1;CharGC;  while(gc<'0'|| Gc>'9') sig= (gc=='-')?-1:1, gc=GetChar ();  while(gc>='0'&&gc<='9') ret=ret*Ten+gc-'0', gc=GetChar (); returnret*Sig;}voidAddintAintb) {to[cnt]=C; NEXT[CNT]=Head[a]; Head[a]=cnt++;}voidDFS1 (intx) {Size[x]=1;  for(inti=head[x];i!=-1; i=Next[i]) {        if(to[i]!=Fa[x]) {Fa[to[i]]=Y; Deep[to[i]]=deep[x]+1;            DFS1 (To[i]); SIZE[X]+=Size[to[i]]; if(Size[to[i]]>size[son[x]]) son[x]=To[i]; }    }}voidDFS2 (intXintTP) {Top[x]=TP; P[X]=++tot; V[P[X]]=U[x]; if(Son[x]) DFS2 (SON[X],TP);  for(inti=head[x];i!=-1; i=Next[i])if(to[i]!=son[x]&&to[i]!=fa[x]) DFS2 (To[i],to[i]);}voidPushup (intx) {S[x]=s[lson]+S[rson];}voidPushdown (intLintRintx) {    intMid=l+r>>1; Tag[lson]+=tag[x],tag[rson]+=Tag[x]; S[lson]+=tag[x]* (Long Long) (mid-l+1), s[rson]+=tag[x]* (Long Long) (Rmid); TAG[X]=0;}voidBuildintLintRintx) {    if(l==R) {S[x]=V[l]; return ; }    intMid=l+r>>1; Build (L,mid,lson), Build (mid+1, R,rson); Pushup (x);}voidUpdata (intLintRintXintAintBintc) {    if(a<=l&&r<=b) {s[x]+=c* (Long Long) (r-l+1), tag[x]+=C; return ;    } pushdown (l,r,x); intMid=l+r>>1; if(b<=mid) Updata (L,MID,LSON,A,B,C); Else if(A>mid) Updata (mid+1, R,rson,a,b,c); ElseUpdata (L,mid,lson,a,b,c), Updata (mid+1, R,rson,a,b,c); Pushup (x);} ll query (intLintRintXintAintb) {    if(a<=l&&r<=b)returnS[x];    Pushdown (L,R,X); intMid=l+r>>1; if(B<=mid)returnquery (L,MID,LSON,A,B); if(A>mid)returnQuery (mid+1, r,rson,a,b); returnQuery (l,mid,lson,a,b) +query (mid+1, r,rson,a,b);}voidChange () {intX=readin (), y=Readin (); Updata (1N1, p[x],p[x],y);}voidAdd () {intX=readin (), y=Readin (); Updata (1N1, p[x],p[x]+size[x]-1, y);//Modify an entire subtrees tree}voidGetans () {intx=Readin (); ll ans=0;  while(top[x]!=1) {ans+=query (1N1, p[top[x]],p[x]); X=Fa[top[x]]; } ans+=query (1N1,1, p[x]); printf ("%lld\n", ans);}intMain () {n=readin (), m=Readin (); memset (Head,-1,sizeof(head)); inti,a,b;  for(i=1; i<=n;i++) u[i]=Readin ();  for(i=1; i<n;i++) {a=readin (), b=Readin ();    Add (A, B), add (B,a); } deep[1]=1; DFS1 (1); DFS2 (1,1); Build (1N1);  for(i=1; i<=m;i++) {a=Readin (); Switch(a) { Case 1: Change (); Break;  Case 2: Add (); Break;  Case 3: Getans (); Break; }    }    return 0;}

"BZOJ4034" [HAOI2015] tree operation tree chain split + segment tree