Codeforces Round #426 (Div. 1), problem: (D) Red-Black Cobweb Solution In C/C++

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#define maxn 100005
#define mod 1000000007
#define ll long long
using namespace std;
ll ksm(ll a, ll b)
{
if(!b) return 1;
ll ns = ksm(a, b >> 1);
ns *= ns, ns %= mod;
if(b & 1) ns *= a, ns %= mod;
return ns;
}
struct edge
{
int u, v, c;
ll w;
bool flag;
edge *rev;
edge *next;
}pool[maxn << 2], *h[maxn];
int cnt = 0;
void addedge(int u, int v, int w, int c)
{
edge *new1 = &pool[cnt++];
new1->u = u, new1->v = v, new1->w = w, new1->c = c, new1->flag = 1, new1->next = h[u], h[u] = new1;
edge *new2 = &pool[cnt++];
new2->u = v, new2->v = u, new2->w = w, new2->c = c, new2->flag = 1, new2->next = h[v], h[v] = new2;
new1->rev = new2, new2->rev = new1;
}
int mxpl = -1, mxs = maxn;
int size[maxn];
void getzx(int a, int fa, int asize)
{
size[a] = 1;
int nmx = 0;
for(edge *p = h[a]; p; p = p->next)
{
if(!p->flag) continue;
if(p->v == fa) continue;
getzx(p->v, a, asize), size[a] += size[p->v], nmx = max(nmx, size[p->v]);
}
int ns = max(nmx, asize – size[a]);
if(ns < mxs) mxs = ns, mxpl = a;
}

int lowbit(int x)
{
return x & (-x);
}
const int pls = maxn * 2;
struct bit1
{
int d[maxn << 2];
int chq[maxn * 20], ed;
bit1()
{
ed = 0;
for(int i = 0; i < (maxn << 2); i++) d[i] = 0;
}
void init()
{
while(ed) d[chq[ed – 1]] = 0, ed–;
}
void ch(int pl, int num)
{
pl += pls;
while(pl < (maxn << 2))
d[pl] += num, chq[ed++] = pl, pl += lowbit(pl);
}
int query(int pl)
{
pl += pls;
int ans = 0;
while(pl)
ans += d[pl], pl -= lowbit(pl);
return ans;
}
}plbit[2];

struct bit2
{
ll d[maxn << 2];
int chq[maxn * 20], ed;
bit2()
{
ed = 0;
for(int i = 0; i < (maxn << 2); i++) d[i] = 1;
}
void init()
{
while(ed) d[chq[ed – 1]] = 1, ed–;
}
void ch(int pl, int num)
{
// cout<<“CHANGE”<<pl<<” “<<num<<endl;
pl += pls;
while(pl < (maxn << 2))
d[pl] *= num, d[pl] %= mod, chq[ed++] = pl, pl += lowbit(pl);
}
ll query(int pl)
{
pl += pls;
ll ans = 1;
while(pl)
ans *= d[pl], ans %= mod, pl -= lowbit(pl);
return ans;
}
}mulbit[2];

int ed = 0, q[maxn][3], num[maxn][2];
ll ans = 1;
void dfs(int a, int fa, ll mt)
{
q[ed][0] = num[a][0], q[ed][1] = num[a][1], q[ed][2] = mt, ed++;
for(edge *p = h[a]; p; p = p->next)
{
if(!p->flag) continue;
if(p->v == fa) continue;
num[p->v][0] = num[a][0], num[p->v][1] = num[a][1], num[p->v][p->c]++;
dfs(p->v, a, mt * p->w % mod);
}
}
ll bk(ll a){return ksm(a, mod – 2);}
void work(int a, int asize)
{
for(int i = 0; i < 2; i++)
plbit[i].init(), mulbit[i].init();
ll nsize = 1, nl = 1;
for(int j = 0; j < 2; j++)
plbit[j].ch(0, 1), mulbit[j].ch(0, 1);
for(edge *p = h[a]; p; p = p->next)
{
if(!p->flag) continue;
ed = 0;
num[p->v][0] = num[p->v][1] = 0, num[p->v][p->c]++;
dfs(p->v, a, p->w);
for(int i = 0; i < ed; i++)
{
ans *= ksm(q[i][2], nsize), ans %= mod;
ans *= nl, ans %= mod;
}
for(int i = 0; i < ed; i++)
nl *= q[i][2], nl %= mod, nsize++;
for(int i = 0; i < ed; i++)
for(int j = 0; j < 2; j++)
ans *= bk(ksm(q[i][2], plbit[j].query(q[i][j] – 2 * q[i][!j] – 1))),
ans %= mod,
ans *= bk(mulbit[j].query(q[i][j] – 2 * q[i][!j] – 1)),
ans %= mod;

for(int i = 0; i < ed; i++)
for(int j = 0; j < 2; j++)
plbit[j].ch(2 * q[i][!j] – q[i][j], 1),
mulbit[j].ch(2 * q[i][!j] – q[i][j], q[i][2]);
}
getzx(a, -1, asize);
for(edge *p = h[a]; p; p = p->next)
{
if(!p->flag) continue;
p->flag = p->rev->flag = 0;
mxpl = -1, mxs = maxn;
getzx(p->v, -1, size[p->v]);
work(mxpl, size[p->v]);
}
}
int main()
{
int n;
scanf(“%d”, &n);
for(int i = 1; i < n; i++)
{
int u, v, w, c;
scanf(“%d%d%d%d”, &u, &v, &w, &c);
addedge(u, v, w, c);
}
work(1, n);
printf(“%I64d\n”, ans);
return 0;
}

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <algorithm>
#define maxn 100005
#define mod 1000000007
#define ll long long
using namespace std;
ll ksm(ll a, ll b)
{
if(!b) return 1;
ll ns = ksm(a, b >> 1);
ns *= ns, ns %= mod;
if(b & 1) ns *= a, ns %= mod;
return ns;
}
struct edge
{
int u, v, c;
ll w;
bool flag;
edge *rev;
edge *next;
}pool[maxn << 2], *h[maxn];
int cnt = 0;
void addedge(int u, int v, int w, int c)
{
edge *new1 = &pool[cnt++];
new1->u = u, new1->v = v, new1->w = w, new1->c = c, new1->flag = 1, new1->next = h[u], h[u] = new1;
edge *new2 = &pool[cnt++];
new2->u = v, new2->v = u, new2->w = w, new2->c = c, new2->flag = 1, new2->next = h[v], h[v] = new2;
new1->rev = new2, new2->rev = new1;
}
int mxpl = -1, mxs = maxn;
int size[maxn];
void getzx(int a, int fa, int asize)
{
size[a] = 1;
int nmx = 0;
for(edge *p = h[a]; p; p = p->next)
{
if(!p->flag) continue;
if(p->v == fa) continue;
getzx(p->v, a, asize), size[a] += size[p->v], nmx = max(nmx, size[p->v]);
}
int ns = max(nmx, asize – size[a]);
if(ns < mxs) mxs = ns, mxpl = a;
}

int lowbit(int x)
{
return x & (-x);
}
const int pls = maxn * 2;
struct bit1
{
int d[maxn << 2];
int chq[maxn * 20], ed;
bit1()
{
ed = 0;
for(int i = 0; i < (maxn << 2); i++) d[i] = 0;
}
void init()
{
while(ed) d[chq[ed – 1]] = 0, ed–;
}
void ch(int pl, int num)
{
pl += pls;
while(pl < (maxn << 2))
d[pl] += num, chq[ed++] = pl, pl += lowbit(pl);
}
int query(int pl)
{
pl += pls;
int ans = 0;
while(pl)
ans += d[pl], pl -= lowbit(pl);
return ans;
}
}plbit[2];

struct bit2
{
ll d[maxn << 2];
int chq[maxn * 20], ed;
bit2()
{
ed = 0;
for(int i = 0; i < (maxn << 2); i++) d[i] = 1;
}
void init()
{
while(ed) d[chq[ed – 1]] = 1, ed–;
}
void ch(int pl, int num)
{
// cout<<“CHANGE”<<pl<<” “<<num<<endl;
pl += pls;
while(pl < (maxn << 2))
d[pl] *= num, d[pl] %= mod, chq[ed++] = pl, pl += lowbit(pl);
}
ll query(int pl)
{
pl += pls;
ll ans = 1;
while(pl)
ans *= d[pl], ans %= mod, pl -= lowbit(pl);
return ans;
}
}mulbit[2];

int ed = 0, q[maxn][3], num[maxn][2];
ll ans = 1;
void dfs(int a, int fa, ll mt)
{
q[ed][0] = num[a][0], q[ed][1] = num[a][1], q[ed][2] = mt, ed++;
for(edge *p = h[a]; p; p = p->next)
{
if(!p->flag) continue;
if(p->v == fa) continue;
num[p->v][0] = num[a][0], num[p->v][1] = num[a][1], num[p->v][p->c]++;
dfs(p->v, a, mt * p->w % mod);
}
}
ll bk(ll a){return ksm(a, mod – 2);}
void work(int a, int asize)
{
for(int i = 0; i < 2; i++)
plbit[i].init(), mulbit[i].init();
ll nsize = 1, nl = 1;
for(int j = 0; j < 2; j++)
plbit[j].ch(0, 1), mulbit[j].ch(0, 1);
for(edge *p = h[a]; p; p = p->next)
{
if(!p->flag) continue;
ed = 0;
num[p->v][0] = num[p->v][1] = 0, num[p->v][p->c]++;
dfs(p->v, a, p->w);
for(int i = 0; i < ed; i++)
{
ans *= ksm(q[i][2], nsize), ans %= mod;
ans *= nl, ans %= mod;
}
for(int i = 0; i < ed; i++)
nl *= q[i][2], nl %= mod, nsize++;
for(int i = 0; i < ed; i++)
for(int j = 0; j < 2; j++)
ans *= bk(ksm(q[i][2], plbit[j].query(q[i][j] – 2 * q[i][!j] – 1))),
ans %= mod,
ans *= bk(mulbit[j].query(q[i][j] – 2 * q[i][!j] – 1)),
ans %= mod;

for(int i = 0; i < ed; i++)
for(int j = 0; j < 2; j++)
plbit[j].ch(2 * q[i][!j] – q[i][j], 1),
mulbit[j].ch(2 * q[i][!j] – q[i][j], q[i][2]);
}
getzx(a, -1, asize);
for(edge *p = h[a]; p; p = p->next)
{
if(!p->flag) continue;
p->flag = p->rev->flag = 0;
mxpl = -1, mxs = maxn;
getzx(p->v, -1, size[p->v]);
work(mxpl, size[p->v]);
}
}
int main()
{
int n;
scanf(“%d”, &n);
for(int i = 1; i < n; i++)
{
int u, v, w, c;
scanf(“%d%d%d%d”, &u, &v, &w, &c);
addedge(u, v, w, c);
}
work(1, n);
printf(“%I64d\n”, ans);
return 0;
}

More from author

LEAVE A REPLY

Please enter your comment!
Please enter your name here

Related posts

Advertismentspot_img

Latest posts

Everything You Need to Know About Viral Kitchen Gadgets Right Now

📢 As an Amazon Associate, I earn from qualifying purchases.There comes a point when you realise that tolerating a frustrating situation is costing you...

Ideas on modernizing the open-source desktop

For decades, the Linux community has delivered unparalleled stability, modularity, and power. However, user experience and interface design often lag behind proprietary alternatives. Veteran...

The Ultimate Viral Kitchen Gadgets Guide You Didnt Know You Needed

📢 As an Amazon Associate, I earn from qualifying purchases.I stumbled across this while desperately trying to fix a recurring problem in my routine....

Want to stay up to date with the latest news?

We would love to hear from you! Please fill in your details and we will stay in touch. It's that simple!