CF154C Double Profiles 题解

合集 - 题解(22)

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16.CF154C Double Profiles 题解2024-04-16

17.[ABC208D] Shortest Path Queries 2 题解2024-04-1718.[ABC212E] Safety Journey 题解2024-04-1719.[ABC229E] Graph Destruction 题解2024-04-1720.[ABC240E] Ranges on Tree 题解2024-04-1821.[ABC261E] Many Operations 题解2024-05-1222.P10842 【MX-J2-T3】Piggy and Trees 题解2024-08-05

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CF154C Double Profiles 题解

思路解析

题目说的很明白，求有多少个无序点对 $(i,j)$，使得与 $i$ 直接相连的点集与直接与 $j$ 相连的点集完全相等。我们想到如果直接判断每个 $i,j$ 肯定会超时，所以我们想把每一个与任意一点直接相连的点集进行压缩，可以想到使用字符串哈希的想法压缩一个点集。如果两点的点集哈希后相等，说明两点的点集也相等。

注意，需要特判点度为 $0$ 的情况，以及需要分别讨论 $i,j$ 之间有连边和没连边的情况，最后注意这题卡 map。

code

#include<bits/stdc++.h>
using namespace std;
#define ull unsigned long long
const ull p = 13;
const int N = 1e6 + 10;
int n, m, sz[N];
ull hs[N], pw[N];
signed main() {
	cin >> n >> m;
	pw[0] = 1; for(int i = 1; i <= n; i++) pw[i] = pw[i - 1] * p;	//预处理
	for(int i = 1, u, v; i <= m; i++) {
		cin >> u >> v;
		sz[u]++; sz[v]++;
		hs[u] += pw[v];	//字符串哈希
		hs[v] += pw[u];
	}
	vector<ull> v, v1;
	long long ans = 0;
	for(int i = 1; i <= n; i++) {
		if(sz[i] > 0) v.push_back(hs[i]);	//记得特判入度为 0
		else v.push_back(0);
		v1.push_back(hs[i] + pw[i]);
	}
	sort(v.begin(), v.end()); sort(v1.begin(), v1.end());
	for(int i = 0, j = i; i < v.size(); i = j) {
		j = i; while(j < v.size() && v[j] == v[i]) j++;
		ans += (long long)(j - i) * (long long)(j - i - 1) / 2;
	}
	for(int i = 0, j = i; i < v1.size(); i = j) {
		j = i; while(j < v1.size() && v1[j] == v1[i]) j++;
		ans += (long long)(j - i) * (long long)(j - i - 1) / 2;
	}
	cout << ans;
	return 0;
}