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【线段树】 HDOJ 5274 Dylans loves tree

2015-06-23 18:48 381 查看
用dfs序构建线段树,然后用lca求出两点间路径的xor和。。。

#include <iostream>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 100005
#define maxm 200005
#define eps 1e-7
#define mod 998244353
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

struct Edge
{
int v;
Edge *next;
}*H[maxn], *edges, E[maxm];

int anc[maxn][20];
int dep[maxn];
int in[maxn];
int out[maxn];
int a[maxn];
int sum[maxn << 4];
int lazy[maxn << 4];
const int M = 20;
int dfs_clock, n, m;

void addedges(int u, int v)
{
edges->v = v;
edges->next = H[u];
H[u] = edges++;
}

void dfs(int u, int fa)
{
dfs_clock++;
in[u] = dfs_clock;
anc[u][0] = fa;
for(Edge *e = H[u]; e; e = e->next) if(e->v != fa) {
int v = e->v;
dep[v] = dep[u] + 1;
dfs(v, u);
}
out[u] = dfs_clock;
}

int to(int u, int d)
{
for(int i = M - 1; i >= 0; i--) if(dep[anc[u][i]] >= d) u = anc[u][i];
return u;
}

int lca(int u, int v)
{
if(dep[u] < dep[v]) swap(u, v);
u = to(u, dep[v]);
for(int i = M - 1; i >= 0; i--) if(anc[u][i] != anc[v][i]) u = anc[u][i], v = anc[v][i];
return u == v ? u : anc[u][0];
}

void init()
{
dfs_clock = 0;
dfs(1, 1);
for(int i = 1; i < M; i++)
for(int j = 1; j <= n; j++)
anc[j][i] = anc[anc[j][i-1]][i-1];
}

void pushup(int o)
{
sum[o] = sum[ls] ^ sum[rs];
}

void pushdown(int o)
{
if(lazy[o]) {
sum[ls] ^= lazy[o];
sum[rs] ^= lazy[o];
lazy[ls] ^= lazy[o];
lazy[rs] ^= lazy[o];
lazy[o] = 0;
}
}

void build(int o, int L, int R)
{
lazy[o] = sum[o] = 0;
if(L == R) return;
int mid = (L + R) >> 1;
build(lson);
build(rson);
pushup(o);
}

void update(int o, int L, int R, int ql, int qr, int v)
{
if(ql <= L && qr >= R) {
lazy[o] ^= v;
sum[o] ^= v;
return;
}
pushdown(o);
int mid = (L + R) >> 1;
if(ql <= mid) update(lson, ql, qr, v);
if(qr > mid) update(rson, ql, qr, v);
pushup(o);
}

int query(int o, int L, int R, int q)
{
if(L == R) return sum[o];
pushdown(o);
int ans = 0, mid = (L + R) >> 1;
if(q <= mid) ans = query(lson, q);
else ans = query(rson, q);
pushup(o);
return ans;
}

void work()
{
int u, v;
scanf("%d%d", &n, &m);
edges = E;
memset(H, 0, sizeof H);
for(int i = 1; i < n; i++) {
scanf("%d%d", &u, &v);
addedges(u, v);
addedges(v, u);
}
for(int i = 1; i <= n; i++) scanf("%d", &a[i]), a[i]++;
init();
build(1, 1, n);
for(int i = 1; i <= n; i++) update(1, 1, n, in[i], out[i], a[i]);

while(m--) {
int kk;
scanf("%d%d%d", &kk, &u, &v);
if(kk) {
int c = lca(u, v);
int ans = query(1, 1, n, in[u]) ^ query(1, 1, n, in[v]) ^ a[c];
if(ans) printf("%d\n", ans-1);
else printf("-1\n");
}
else {
++v;
update(1, 1, n, in[u], out[u], a[u] ^ v);
a[u] = v;
}
}
}

int main()
{
int _;
scanf("%d", &_);
while(_--) work();

return 0;
}
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