Codeforces-1323E Instant Noodles
Codeforces-1323E Instant Noodles
Wu got hungry after an intense training session, and came to a nearby store to buy his favourite instant noodles. After Wu paid for his purchase, the cashier gave him an interesting task.
You are given a bipartite graph with positive integers in all vertices of the right half. For a subset \(S\) of vertices of the left half we define \(N(S)\) as the set of all vertices of the right half adjacent to at least one vertex in \(S\), and \(f(S)\) as the sum of all numbers in vertices of \(N(S)\). Find the greatest common divisor of \(f(S)\) for all possible non-empty subsets \(S\) (assume that GCD of empty set is \(0\)).
Wu is too tired after his training to solve this problem. Help him!
Input
The first line contains a single integer \(t\) (\(1 \leq t \leq 500\,000\)) — the number of test cases in the given test set. Test case descriptions follow.
The first line of each case description contains two integers \(n\) and \(m\) (\(1~\leq~n,~m~\leq~500\,000\)) — the number of vertices in either half of the graph, and the number of edges respectively.
The second line contains \(n\) integers \(c_i\) (\(1 \leq c_i \leq 10^{12}\)). The \(i\)-th number describes the integer in the vertex \(i\) of the right half of the graph.
Each of the following \(m\) lines contains a pair of integers \(u_i\) and \(v_i\) (\(1 \leq u_i, v_i \leq n\)), describing an edge between the vertex \(u_i\) of the left half and the vertex \(v_i\) of the right half. It is guaranteed that the graph does not contain multiple edges.
Test case descriptions are separated with empty lines. The total value of \(n\) across all test cases does not exceed \(500\,000\), and the total value of \(m\) across all test cases does not exceed \(500\,000\) as well.
Output
For each test case print a single integer — the required greatest common divisor.
Example
Input
3
2 4
1 1
1 1
1 2
2 1
2 2
3 4
1 1 1
1 1
1 2
2 2
2 3
4 7
36 31 96 29
1 2
1 3
1 4
2 2
2 4
3 1
4 3
Output
2
1
12
Note
The greatest common divisor of a set of integers is the largest integer \(g\) such that all elements of the set are divisible by \(g\).
In the first sample case vertices of the left half and vertices of the right half are pairwise connected, and \(f(S)\) for any non-empty subset is \(2\), thus the greatest common divisor of these values if also equal to \(2\).
In the second sample case the subset \(\{1\}\) in the left half is connected to vertices \(\{1, 2\}\) of the right half, with the sum of numbers equal to \(2\), and the subset \(\{1, 2\}\) in the left half is connected to vertices \(\{1, 2, 3\}\) of the right half, with the sum of numbers equal to \(3\). Thus, \(f(\{1\}) = 2\), \(f(\{1, 2\}) = 3\), which means that the greatest common divisor of all values of \(f(S)\) is \(1\).
题意
给定一个二分图,每一个右半部分的点有一个点权,对于每一个左边的点集\(S\),有一个与之相连的右边的点集\(N(S)\),其点权和为\(f(S)\),问对于所有\(S\),\(f(S)\)的最大公约数为多少
题解
首先\(gcd(a,b)=gcd(a+b,b)\),所以我们只需要考虑那些没有交集最基本的右边集合即可.
我们要求的是所有\(f(S)\),所以我们考虑右边点的集合,对于一个右边的,其连向的左边的点集为\(S\),如果有两个右边的点连向的点集均为\(S\),则合并这两个点,因为他们必定同属于一个集合,对于所有不重复的右边基本集合均算出后做一遍gcd即可.
复杂度:对于每条边都会被扫一遍,记录对应的集合,加上排序,map处理映射,复杂度\(O(mlogm)\)
如何看右边的点连向的S是否相同,可以使用hash,在搜题解时发现了新的hash方法,一个是直接使用map映射vector,另一个是用异或.
- 使用map映射vector
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 50;
typedef long long ll;
ll c[N];
ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);}
int main() {
int t; scanf("%d", &t);
while (t--) {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) scanf("%lld", &c[i]);
vector<int> a[n + 1];
for (int i = 1; i <= m; i++) {
int u, v;
scanf("%d%d", &u, &v);
a[v].push_back(u);
}
for (int i = 1; i <= n; i++) {
if (a[i].size()) {
sort(a[i].begin(), a[i].end());
}
}
map<vector<int>, ll> mp;
for (int i = 1; i <= n; i++) {
if (!a[i].size()) continue;
if (!mp[a[i]]) {
mp[a[i]] = c[i];
}
else mp[a[i]] += c[i];
}
ll ans = 0;
for (auto e : mp) ans = gcd(ans, e.second);
printf("%lld\n", ans);
}
return 0;
}
//795ms
-
使用异或作为hash值
这里学习了https://blog.csdn.net/iamhpp/article/details/104826481的做法
就是给每个位置生成大随机数，然后将一个集合的异或起来，作为哈希值。
然后再用个哈希表把相同哈希值对应的 \(c_i\)加起来。
可以用 mt19937_64 生成大随机数
这个复杂度是最小的，接近 O(n)这样就不需要排序了,map的常数也变小了很多,注意mt19937返回值为unsigned
#include <bits/stdc++.h>
using namespace std;
const int N = 5e5 + 50;
typedef long long ll;
typedef unsigned long long ull;
ll c[N];
ll gcd(ll a, ll b) {return b == 0 ? a : gcd(b, a % b);}
mt19937_64 rdn(time(0));
ull h[N];
ull rd[N];
int main() {
int t; scanf("%d", &t);
while (t--) {
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++) {
rd[i] = rdn();
h[i] = 0;
scanf("%lld", &c[i]);
}
for (int i = 1; i <= m; i++) {
int u, v;
scanf("%d%d", &u, &v);
h[v] ^= rd[u];
}
unordered_map<ull, ll> mp;
for (int i = 1; i <= n; i++) {
if (!h[i]) continue;
mp[h[i]] += c[i];
}
ll ans = 0;
for (auto e : mp) ans = gcd(ans, e.second);
printf("%lld\n", ans);
}
return 0;
}
//561ms
