題目描述

You ye Jiu yuan is the daughter of the Great GOD Emancipator.? And when she becomes an adult, she will be queen of Tusikur, so she wanted to travel the world while she was still young. In a country, she found a small pub called Whitehouse. Just as she was about to go in for a drink, the boss Carola appeared. And ask her to solve this problem or she will not be allowed to enter the pub. The problem description is as follows:
There is a tree with n nodes, each node i contains weight a[i], the initial value of a[i] is 0.? The root number of the tree is 1. Now you need to do the following operations:
1) Multiply all weight on the path from u to v by x
2) For all weight on the path from u to v, increasing x to them
3) For all weight on the path from u to v, change them to the bitwise NOT of them
4) Ask the sum of the weight on the path from u to v
The answer modulo 2^64.

Jiu Yuan is a clever girl, but she was not good at algorithm, so she hopes that you can help her solve this problem. Ding~~~

The bitwise NOT is a unary operation that performs logical negation on each bit, forming the ones' complement of the given binary value. Bits that are 0 become 1, and those that are 1 become 0. For example:

NOT 0111 (decimal 7) = 1000 (decimal 8)
NOT 10101011 = 01010100

輸入

The input contains multiple groups of data.
For each group of data, the first line contains a number of n, and the number of nodes.
The second line contains (n - 1) integers bi, which means that the father node of node (i +1) is bi.
The third line contains one integer m, which means the number of operations，
The next m lines contain the following four operations:
At first, we input one integer opt
1) If opt is 1, then input 3 integers, u, v, x, which means multiply all weight on the path from u to v by x
2) If opt is 2, then input 3 integers, u, v, x, which means for all weight on the path from u to v, increasing x to them
3) If opt is 3, then input 2 integers, u, v, which means for all weight on the path from u to v, change them to the bitwise NOT of them
4) If opt is 4, then input 2 integers, u, v, and ask the sum of the weights on the path from u to v

1 ≤ n, m, u, v ≤ 10^5
1 ≤ x < 2^64

輸出

For each operation 4, output the answer.

樣例輸入

7 1 1 1 2 2 4 5 2 5 6 1 1 1 6 2 4 5 6 3 5 2 4 2 2 2 1 4 3 1 2 4 1 2 3 1 1 4 1 1

樣例輸出

5 18446744073709551613 18446744073709551614 0

題意 給一棵n個節點的有根樹，每個節點有權值，初始是0，m次操作 1 u v x：給u v路徑上的點權值*x 2 u v x：給u v路徑上的點權值+x 3 u v：給u v路徑上的點權值取反 4 u v：詢問u v路徑上的權值和，對2^64取模樹鏈剖分：https://wenku.baidu.com/view/a088de01eff9aef8941e06c3.html 然后如果沒有取反操作，線段樹維護和sum，加法標記add和乘法標記mul即可 對于取反操作，因為是對2^64取模的，即x+(!x)=2^64-1，所以x=(2^64-1)-x，因此取反就變成乘法和加法了：!x=(-1)*x+(-1) （-1對于2^64取模后是（2^64-1)） #include <bits/stdc++.h> #define ull unsigned long long using namespace std; const int N=1e5+100; int n,m,tot,cnt; int fa[N],last[N]; int son[N],deep[N],dfn[N],num[N],top[N];//重兒子 深度 dfs序 子樹規模 所在重鏈的頂端節點 ull sum[N*4],add[N*4],mul[N*4]; struct orz{int v,nex;}e[N]; void init() {cnt=0;tot=0;memset(last,0,sizeof(last));memset(son,-1,sizeof(son)); } void Inses(int x,int y) {cnt++;e[cnt].v=y;e[cnt].nex=last[x];last[x]=cnt; } void dfs1(int x,int d) {deep[x]=d;num[x]=1;for (int i=last[x];i;i=e[i].nex){int v=e[i].v;dfs1(v,d+1);num[x]+=num[v];if (son[x]==-1 || num[v]>num[son[x]]) son[x]=v;} } void dfs2(int x,int sp) {top[x]=sp;dfn[x]=++tot;if (son[x]==-1) return ;dfs2(son[x],sp);for (int i=last[x];i;i=e[i].nex){int v=e[i].v;if (v!=son[x]) dfs2(v,v);} } void PushUp(int s) {sum[s]=sum[s<<1]+sum[s<<1|1]; } void PushDown(int s,int l,int r) {if (mul[s]!=1){mul[s<<1]*=mul[s];mul[s<<1|1]*=mul[s];add[s<<1]*=mul[s];add[s<<1|1]*=mul[s];sum[s<<1]*=mul[s];sum[s<<1|1]*=mul[s];mul[s]=1;}if (add[s]){add[s<<1]+=add[s];add[s<<1|1]+=add[s];int mid=(l+r)>>1;sum[s<<1]+=(ull)(mid-l+1)*add[s];sum[s<<1|1]+=(ull)(r-mid)*add[s];add[s]=0;} }void build(int s,int l,int r) {sum[s]=add[s]=0;mul[s]=1;if (l==r) return ;int m=(l+r)>>1;build(s<<1,l,m); build(s<<1|1,m+1,r);PushUp(s); } void update(int s,int l,int r,int L,int R,ull val,int op) {//printf("s=%d,l=%d,r=%d,L=%d,R=%d\n",s,l,r,L,R);if (L<=l&&r<=R){if (l!=r) PushDown(s,l,r);if (op==1){mul[s]*=val;add[s]*=val;sum[s]*=val;}else if (op==2){add[s]+=val;sum[s]+=(ull)(r-l+1)*val;}else{mul[s]*=val;add[s]*=val;add[s]+=val;sum[s]=(ull)(r-l+1)*val-sum[s];}return;}PushDown(s,l,r);int mid=(l+r)>>1;if (L<=mid) update(s<<1,l,mid,L,R,val,op);if (R>mid) update(s<<1|1,mid+1,r,L,R,val,op);PushUp(s); } ull query(int s,int l,int r,int L,int R) {if (L<=l&&r<=R) return sum[s];PushDown(s,l,r);int mid=(l+r)>>1;ull ans=0;if (L<=mid) ans+=query(s<<1,l,mid,L,R);if (R>mid) ans+=query(s<<1|1,mid+1,r,L,R);PushUp(s);return ans; } void solve(int op,int x, int y,ull val) {if (op==3) val=-1;if (op<=3){while (top[x]!=top[y]){if (deep[top[x]]<deep[top[y]]) swap(x, y);update(1,1,n,dfn[top[x]],dfn[x],val,op);x=fa[top[x]];}if (deep[x]>deep[y]) swap(x,y);update(1,1,n,dfn[x],dfn[y],val,op);}else{ull ans=0;while (top[x]!=top[y]){if (deep[top[x]]<deep[top[y]]) swap(x, y);ans+=query(1,1,n,dfn[top[x]],dfn[x]);x=fa[top[x]];}if (deep[x]>deep[y]) swap(x,y);ans+=query(1,1,n,dfn[x],dfn[y]);printf("%llu\n",ans);} }int main() {while (scanf("%d",&n)!=EOF){init();for (int i=2;i<=n;i++){scanf("%d",&fa[i]);Inses(fa[i],i);}dfs1(1,0);dfs2(1,1);build(1,1,n);scanf("%d",&m);int op,u,v; ull x;while (m--){scanf("%d",&op);if (op==1 || op==2) scanf("%d%d%llu",&u,&v,&x);else scanf("%d%d",&u,&v);solve(op,u,v,x);}}return 0; } View Code ?

轉載于:https://www.cnblogs.com/tetew/p/11293766.html

總結

以上是生活随笔為你收集整理的2018焦作网络赛-E- Jiu Yuan Wants to Eat的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。