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**HackerEarth City and Fireman Vincent problem solution**Fatland is a town with N cities numbered 1, 2, ..., N, connected with 2-way roads. Vincent is a villian who wants to set the entire town ablaze. For each city, there is a certain risk factor E[i] for setting that city ablaze. But once a city c is set ablaze, all cities that can be reached from c will be set ablaze immediately.The overall risk factor of setting some number of cities ablaze is the sum of the risk factors of setting each of the cities ablaze. Vincent's goal is to set the whole town ablaze with minimal risk by picking some set of cities to set ablaze. Let M be the minimum risk he takes when setting the entire city ablaze. Calculate the number of ways to pick a set of cities to obtain a risk of M.

## HackerEarth City and Fireman Vincent problem solution.

`count = 0`

best = 10000000000000

g = [[] for _ in range(1005)]

v = [0] * 1005

def dfs(x):

global best

global count

if best == e[x]:

count += 1

if best > e[x]:

best = e[x]

count = 1

v[x] = 1

for i in g[x]:

if v[i] == 0:

dfs(i)

n = int(input())

e = [0]+list(map(int, input().split(' ')))

k = int(input())

for i in range(k):

x, y = map(int, input().split(' '))

g[x].append(y)

g[y].append(x)

ans = 1

for i in range(1, n+1):

best = 10000000000000

count = 0

if (not v[i]):

dfs(i)

ans = ans * count

print(ans%(10**9+7))

### Second solution

`#include <bits/stdc++.h>`

using namespace std;

typedef long long LL;

const int lim = 1e3 + 1;

const int maxn = 1e3;

const int maxk = 1e3;

const int LL mod = 1e9 + 7;

vector<LL> E(lim);

vector<int> parent(lim);

int ancestor(int a)

{

if (a != parent[a])

parent[a] = ancestor(parent[a]);

return parent[a];

}

void merge(int a, int b)

{

a = ancestor(a);

b = ancestor(b);

if (a != b)

parent[a] = b;

}

int main()

{

int n;

scanf("%d", &n);

assert(1 <= n and n <= maxn);

for (int i = 1; i <= n; i++)

scanf("%lld", &E[i]);

for (int i = 1; i <= n; i++)

parent[i] = i;

int k;

scanf("%d", &k);

assert(1 <= k and k <= maxk);

for (int i = 1; i <= k; i++)

{

int city1, city2;

scanf("%d%d", &city1, &city2);

merge(city1, city2);

}

for (int i = 1; i <= n; i++)

parent[i] = ancestor(i);

map<int, multiset<int> > mp;

for (int i = 1; i <= n; i++)

mp[parent[i]].insert(E[i]);

LL ans = 1;

for (auto foo: mp)

{

LL cnt = 1;

int low = *(foo.second.begin());

foo.second.erase(foo.second.begin());

for (auto bar: foo.second)

{

if (bar != low)

break;

cnt++;

}

ans = (ans * cnt) % mod;

}

printf("%lld\n", ans);

return 0;

}

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