題目鏈接
A tree is a connected graph without cycles. You are given a rooted tree with nn nodes, labeled from 1 to n1ton. The tree is rooted at node 11. The parent of the ii-th node is f_{a_i}fai. The edge weight between the ii-th node and f_{a_i}fai is w_iwi
- E(u, v)E(u,v) is a set of all the nodes through a path which is from uu to vv, including uu and vv. For example,
E(5,3) = \{5, 2, 1, 3\}, E(4,6) = \{4, 3, 6\}, E(2,5) = \{2,5\}E(5,3)={5,2,1,3},E(4,6)={4,3,6},E(2,5)={2,5}
- X(u, v)X(u,v) represents the XOR of all edge weights on the path from uu to vv
For example, in the case given above,
X(1,5) = 1X(1,5)=1 xorxor 1=01=0,X(4,6) = 1X(4,6)=1 xorxor 3=23=2,X(2,5) = 1X(2,5)=1,X(3,5) = 2X(3,5)=2 xorxor 11 xorxor 11 =2=2You need to calculate the answer to the following formula:
\displaystyle\sum_{u=1}^n \displaystyle\sum_{v=1}^n \displaystyle\sum_{u' \in E(u,v)} \displaystyle\sum_{v' \in E(u,v)} [u < v][ u' < v'][X(u',v')=0]u=1∑nv=1∑nu′∈E(u,v)∑v′∈E(u,v)∑[u<v][u′<v′][X(u′,v′)=0]
The answer modulo 10000000071000000007
- XOR is equivalent to '^' in c / c++ / java / python. If both bits in the compared position of the bit patterns are 00 or 11, the bit in the resulting bit pattern is 00, otherwise 11.
Input
Each test file contains a single test case. In each test file:
The first line contains one integer n(1 \le n \le 10^5)n(1≤n≤105) — the number of nodes in the tree.
Then n - 1n−1 lines follow. ii-th line contains two integers f_{a_i}(1 \le f_{a_i} < i)fai(1≤fai<i), w_i(0 \le w_i \le 10^{18})wi(0≤wi≤1018) —The parent of the ii-th node and the edge weight between the ii-th node and f_{a_i} (ifai(i start from 2)2).
Output
Print a single integer — the answer of this problem, modulo 1000000007.
題意是這樣的：一個由N個點組成的樹，我們要詢問是這樣的，假如有一條鏈的異或值爲0，也就是代表着所有該鏈上的邊的值異或起來爲0，然後求所有包含這條鏈的鏈的個數，當然，可以重複出現鏈。
主要就是求這樣的一個東西。
思路：當然，我們這裏先不講樹形dp的解法（當然是因爲我還沒有解出來的緣故），先講講點分治。
首先的話，就是要避免重複，我們將一顆樹不斷的向下分治，首先就必須要滿足分治過後的答案是可以合成起來的。在這裏，我們從一個點（子樹的重心）出發，以它爲核心開始分割，然後就變成了我們現在訪問到的點到已有點的答案之和了，不存在重複，所以就無需減去什麼了。
只是這裏有個重點需要處理的內容，就是假如從該點出發，異或到某個點的時候值爲0了，這時候需要把這個答案加上，因爲當我們分治的時候，重心的那個點是沒有計算在內的，假如有它的答案，依然是要算進去的。
還有就是，假如從一個點開始分治的時候，兩邊的子樹某到該點的異或值爲0，我們不僅要算上述的答案還是要去算在已有子樹中的爲0的答案求和。
最後，別忘了要去求一個mod。
給一組測試樣例，方便理解（囊括了上面基本的內容）：
10
1 1
1 1
2 2
3 2
4 1
4 1
5 1
5 2
8 1
ans:58
My Code：
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <unordered_map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f
#define efs 1e-6
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
using namespace std;
typedef double lb;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 1e5 + 7;
const ll mod = 1e9 + 7;
int pre[maxN], head[maxN], cnt, rt;
bool vis[maxN];
ll num_son[maxN], son[maxN], siz[maxN], dis[maxN], ans, N, all, mx;
unordered_map<ll, ll> mp;
struct Eddge
{
int nex, to; ll val;
Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), val(c) {}
}edge[maxN<<1];
inline void addEddge(int u, int v, ll w)
{
edge[cnt] = Eddge(head[u], v, w);
head[u] = cnt++;
}
inline void _add(int u, int v, ll w) { addEddge(u, v, w); addEddge(v, u, w); }
void pre_dfs(int u, int fa)
{
pre[u] = fa;
siz[u] = 1;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == fa) continue;
pre_dfs(v, u);
siz[u] += siz[v];
}
}
int fath[maxN];
void findroot(int u, int fa)
{
num_son[u] = 1; son[u] = 0;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(vis[v] || v == fa) continue;
findroot(v, u);
num_son[u] += num_son[v];
if(num_son[v] > son[u]) son[u] = num_son[v];
}
son[u] = max(son[u], all - num_son[u]);
if(son[u] < mx) { mx = son[u]; rt = u; }
}
struct node
{
int id; ll val;
node(int a=0, ll b=0):id(a), val(b) {}
}Stap[maxN];
int Stop;
void get_dis(int u, int fa)
{
Stap[++Stop] = node(u, dis[u]);
fath[u] = fa;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(vis[v] || v == fa) continue;
dis[v] = dis[u] ^ edge[i].val;
get_dis(v, u);
}
}
void solve(int u)
{
mp.clear();
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(vis[v]) continue;
Stop = 0;
dis[v] = edge[i].val;
get_dis(v, u);
for(int j=1, id; j<=Stop; j++)
{
id = Stap[j].id;
if(fath[id] == pre[id])
{
ans = (ans + mp[Stap[j].val] * siz[id]) % mod;
if(!Stap[j].val)
{
if(pre[v] == fath[v]) ans = (ans + siz[id] * (N - siz[v])) % mod;
else ans = (ans + siz[id] * siz[u]) % mod;
}
}
else
{
ans = (ans + mp[Stap[j].val] * (N - siz[fath[id]])) % mod;
if(!Stap[j].val)
{
if(pre[v] == fath[v]) ans = (ans + ((ll)N - siz[fath[id]]) * (N - siz[v])) % mod;
else ans = (ans + (N - siz[fath[id]]) * siz[u]) % mod;
}
}
}
for(int j=1, id; j<=Stop; j++)
{
id = Stap[j].id;
if(fath[id] == pre[id])
{
mp[Stap[j].val] = (mp[Stap[j].val] + siz[id]) % mod;
}
else
{
mp[Stap[j].val] = (mp[Stap[j].val] + (N - siz[fath[id]])) % mod;
}
}
}
}
void divide(int u)
{
vis[u] = true;
solve(u);
ll totsiz = all;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(vis[v]) continue;
mx = INF; rt = 0;
all = num_son[v] > num_son[u] ? totsiz - num_son[u] : num_son[v];
findroot(v, 0);
divide(rt);
}
}
inline void init()
{
cnt = 0; ans = 0; all = N; mx = INF;
for(int i=1; i<=N; i++) head[i] = -1;
for(int i=1; i<=N; i++) vis[i] = false;
}
int main()
{
scanf("%lld", &N);
init();
for(int i=2, uu; i<=N; i++)
{
ll ww;
scanf("%d%lld", &uu, &ww);
_add(uu, i, ww);
}
pre_dfs(1, 0);
findroot(1, 0);
divide(rt);
printf("%lld\n", ans);
return 0;
}