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【树分治】 HDOJ 4812 D Tree

2015-05-27 20:39 260 查看
比较浅显的树分治。。。对于根节点,找出两条路径使得乘积为k就行了。。。。但是这道题卡时间。。。要写的很好才行。。。

#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 400005
#define eps 1e-12
#define mod 1000003
#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
#pragma comment(linker, "/STACK:102400000,102400000")
#define pii pair<int, int>
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;
}E[maxm], *H[maxn], *edges;

pii h[mod + 5];
vector<pair<int, int> > dis;
int size[maxn];
bool done[maxn];
int mx[maxn];
int inv[mod + 5];
int a[maxn];
int kk, T;
pii ans;
int n, root, nsize;

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

void init()
{
T = 0;
edges = E;
memset(H, 0, sizeof H);
memset(done, 0, sizeof done);
memset(h, 0, sizeof h);
}

void getroot(int u, int fa)
{
size[u] = 1, mx[u] = 0;
for(Edge *e = H[u]; e; e = e->next) if(!done[e->v] && e->v != fa) {
int v = e->v;
getroot(v, u);
size[u] += size[v];
mx[u] = max(mx[u], size[v]);
}
mx[u] = max(mx[u], nsize - size[u]);
if(mx[u] < mx[root]) root = u;
}

void getdis(int u, int fa, LL dist)
{
dist = dist * a[u] % mod;
dis.push_back(mp(dist, u));
int t = (LL)kk * inv[dist] % mod;
if(h[t].second == T) {
ans = min(ans, mp(u, h[t].first));
ans = min(ans, mp(h[t].first, u));
}
for(Edge *e = H[u]; e; e = e->next) if(!done[e->v] && e->v != fa) getdis(e->v, u, dist);
}

void solve(int u)
{
T++;
kk = (LL)kk * a[u] % mod;
h[a[u]] = mp(u, T);
done[u] = true;
for(Edge *e = H[u]; e; e = e->next) if(!done[e->v]) {
int v = e->v;
dis.clear();
getdis(v, u, a[u]);
for(int i = 0; i < dis.size(); i++) {
if(h[dis[i].first].second == T) h[dis[i].first] = mp(min(h[dis[i].first].first, dis[i].second), T);
else h[dis[i].first] = mp(dis[i].second, T);
}
}
kk = (LL)kk * inv[a[u]] % mod;
for(Edge *e = H[u]; e; e = e->next) if(!done[e->v]) {
int v = e->v;
nsize = mx[0] = size[v];
getroot(v, root = 0);
solve(root);
}
}

void work()
{
int u, v;
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
for(int i = 1; i < n; i++) {
scanf("%d%d", &u, &v);
addedges(u, v);
addedges(v, u);
}
ans = mp(INF, INF);
nsize = mx[0] = n;
getroot(1, root = 0);
solve(root);
if(ans == mp(INF, INF)) printf("No solution\n");
else printf("%d %d\n", ans.first, ans.second);
}

int main()
{
for(int i = 1; i < mod; i++) inv[i] = powmod(i, mod - 2);
while(scanf("%d%d", &n, &kk)!=EOF) {
init();
work();
}

return 0;
}
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