Dylans loves tree
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 747 Accepted Submission(s): 144
Problem Description
Dylans is given a tree with nodes.
All nodes have a value .Nodes
on tree is numbered by .
Then he is given questions
like that:
①:change
node value
to
②:For
all the value in the path from to ,do
they all appear even times?
For each ② question,it guarantees that there is at most one value that appears odd times on the path.
,
the value and
Input
In the first line there is a test number .
( and
there is at most one testcase that )
For each testcase:
In the first line there are two numbers and .
Then in the next lines
there are pairs of that
stand for a road from to .
Then in the next line there are numbers stand
for value.
In the next lines
there are three numbers.
Output
For each question ② in each testcase,if the value all appear even times output "-1",otherwise output the value that appears odd times.
Sample Input
1 3 2 1 2 2 3 1 1 1 1 1 2 1 1 3
Sample Output
-1 1 Hint If you want to hack someone,N and Q in your testdata must smaller than 10000,and you shouldn‘t print any space in each end of the line.
Source
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先考虑无改动的情况,令Xor[i]表示i到根节点路径上的异或和。则随意节点的(u,v)的异或和能够转化为Xor[u]^Xor[v]^a[LCA(u,v)].考虑改动的情况。改动节点u,仅仅会以u为根的子树的Xor值产生影响,由于一颗子树的dfs序是连续的我们非常自然的想到用线段树去维护他,pSeg[u]表示u在dfs序中的位置,siz[u]表示以u为根的子树大小,则这课颗子树相应的区间就是[pSeg[u],pSeg[u]+siz[u]-1],改动的时候仅仅须要将这段区间先异或上原来的值a[u],在异或上要变成的值y,然后改动a[u]
= y;两次异或能够一步到位。直接异或上a[u]^y即可。
#pragma comment(linker, "/STACK:102400000,102400000") #include <iostream> #include <cstdio> #include <cstring> using namespace std; typedef long long ll; const ll mod = 1e9 + 7; const int maxn = 1e5 + 10; #define to first #define next second #define foreach(it,v) for(__typeof(v.begin()) it = v.begin(); it != v.end(); ++it) int pos[maxn],d[20][maxn<<1],wid[maxn<<1],head[maxn]; int a[maxn],depth[maxn],sid,pSeg[maxn],siz[maxn],Xor[maxn]; typedef pair<int,int> Edge; Edge edges[maxn<<1]; int tot = 0,e = 0; void AddEdge(int u,int v) { edges[++e] = make_pair(v,head[u]);head[u] = e; edges[++e] = make_pair(u,head[v]);head[v] = e; } void pre(int u,int fa,int dep = 0,int Xo = 0) { Xo ^= a[u]; Xor[++sid] = Xo; pSeg[u] = sid; siz[u] = 1; d[0][++tot] = u; if(!pos[u]) { pos[u] = tot; depth[u] = dep; } for(int i = head[u]; i ; i = edges[i].next) { int v = edges[i].to; if(v == fa) continue; pre(v,u,dep+1,Xo); siz[u] += siz[v]; d[0][++tot] = u; } } void RMQ_init(int n) { for(int i = 1,w = 1; i <= n; i++) { if((1<<w)<=i) w++; wid[i] = w - 1; } for(int i = 1; (1<<i) <= n; i++) { for(int j = 1; j + (1<<i) - 1 <= n; j++) { d[i][j] = depth[d[i-1][j]] < depth[d[i-1][j+(1<<(i-1))]] ? d[i-1][j] : d[i-1][j+(1<<(i-1))]; } } } int LCA(int u,int v) { u = pos[u]; v = pos[v]; if(u > v) swap(u,v); int k = wid[v-u+1]; return depth[d[k][u]] < depth[d[k][v-(1<<k)+1]] ? d[k][u] : d[k][v-(1<<k)+1]; } int seg[maxn<<2]; int ql,qr,x; void push_down(int o) { seg[o<<1] ^= seg[o]; seg[o<<1|1] ^= seg[o]; seg[o] = 0; } void Modify(int o,int L,int R) { if(ql<=L&&qr>=R) { seg[o] ^= x; return ; } push_down(o); int mid = (L+R)>>1; if(ql<=mid) Modify(o<<1,L,mid); if(qr>mid) Modify(o<<1|1,mid+1,R); } int Query(int o,int L,int R) { if(L == R) { return Xor[L] ^ seg[o]; } int mid = (L+R) >>1; push_down(o); if(x<=mid)return Query(o<<1,L,mid); return Query(o<<1|1,mid+1,R); } int main(int argc, char const *argv[]) { int T;scanf("%d",&T); while(T--) { int N,Q;scanf("%d%d",&N,&Q); e = sid = tot = 0; memset(head,0,sizeof(head[0])*(N+1)); for(int i = 1; i < N; i++) { int u,v;scanf("%d%d",&u,&v); AddEdge(u,v); } for(int i = 1; i <= N; i++) { scanf("%d",a+i); ++a[i]; } pre(1,-1); RMQ_init(tot); memset(seg,0,sizeof(seg[0])*(2*N+10)); while(Q--) { scanf("%d%d%d",&x,&ql,&qr); if(x==0) { qr++; int L = pSeg[ql], R = pSeg[ql] + siz[ql] - 1; x = a[ql] ^ qr; a[ql] = qr; ql = L,qr = R; Modify(1,1,N); }else { x = pSeg[ql]; int ans = Query(1,1,N); x = pSeg[qr]; ans ^= Query(1,1,N); ans ^= a[LCA(ql,qr)]; if(ans==0)puts("-1"); else printf("%d\n", ans-1); } } } return 0; }