AtCoder ABC157D 题解

D - Friend Suggestions

Time Limit: $2$ sec / Memory Limit: $1024$ MB / Score: $400$ points

题意

An SNS has $N$ users - User $1$, User $2$, ⋯, User $N$.
Between these $N$ users, there are some relationships - $M$ friendships and $K$ blockships.
For each $i=1,2,\cdots ,M$, there is a bidirectional friendship between User $A_i$ and User $B_i$.
For each $i=1,2,\cdots ,K$, there is a bidirectional blockship between User $C_i$ and User $D_i$.
We define User $a$ to be a friend candidate for User $b$ when all of the following four conditions are satisfied:

• $a\ne b$.
• There is not a friendship between User $a$ and User $b$.
• There is not a blockship between User $a$ and User $b$.
• There exists a sequence $c_0,c_1,c_2,...,c_L$ consisting of integers between $1$ and $N$ (inclusive) such that $c_0=a$, $c_L=b$, and there is a friendship between User $c_i$ and c_i+1 for each $i=0,1,\cdots ,L-1$.

For each user $i=1,2,...,N$, how many friend candidates does it have?

约束

• 输入中所有数都是整数。
• $2\le N\le 10^5$
• $0\le M\le 10^5$
• $0\le K\le 10^5$
• $1\le A_i,B_i\le N$
• $A_i\ne B_i$
• $1\le C_i,D_i\le N$
• $C_i\ne D_i$
• $(A_i,B_i)\ne (A_j,B_j)(i\ne j)$
• $(A_i,B_i)\ne (B_j,A_j)$
• $(C_i,D_i)\ne (D_j,C_j)(i\ne j)$
• $(A_i,B_i)\ne (C_j,D_j)$
• $(A_i,B_i)\ne (D_j,C_j)$

Input

Input is given from Standard Input in the following format:
$NMK$
$A_1{B}_1$
$A_2{B}_2$
$:$
$A_M{B}_M$
$C_1{D}_1$
$C_2{D}_2$
$:$
$C_K{D}_K$

Output

Print the answers in order, with space in between.

Sample Input 1

4 4 1
2 1
1 3
3 2
3 4
4 1

Sample Output 1

0 1 0 1

There is a friendship between User $2$ and $3$, and between $3$ and $4$. Also, there is no friendship or blockship between User $2$ and $4$. Thus, User $4$ is a friend candidate for User $2$.

However, neither User $1$ or $3$ is a friend candidate for User $2$, so User $2$ has one friend candidate.

Sample Input 2

5 10 0
1 2
1 3
1 4
1 5
3 2
2 4
2 5
4 3
5 3
4 5

Sample Output 2

0 0 0 0 0

Everyone is a friend of everyone else and has no friend candidate.

Sample Input 3

10 9 3
10 1
6 7
8 2
2 5
8 4
7 3
10 9
6 4
5 8
2 6
7 5
3 1

Sample Output 3

1 3 5 4 3 3 3 3 1 0

思路

把整个朋友圈拆分为一个个连通子图，再对于每个用户用对应子图节点数-1(本身)-子图中的跟当前用户有friendship和blockship的逐个得到结果

代码

代码也在这里
注意：只能使用C++>=11的编译器才能通过编译。（我提交时用的是 C++14 (GCC 5.4.1) 编译器）

#include <cstdio>
#include <vector>
#include <queue>
#define maxn 100005
using namespace std;

typedef vector<int>* Graph;

int n, m, k, graphnum[maxn], subgraphsize[maxn];
vector<int> friends[maxn], block[maxn];

template <typename T>
void bfs(const Graph& G, int v, T visit, bool* visited)
{
    
    
	queue<int> q;
	q.push(v);
	while(!q.empty())
	{
    
    
		int x = q.front(); q.pop();
		if(visited[x]) continue;
		visited[x] = true;
		visit(x);
		for(int it: G[x]) q.push(it);
	}
}

void splitgraph(const Graph& G) // Returns: number of subgraphs
{
    
    
	int cnt = 0;
	auto vis = [&cnt](int v) {
    
    
		subgraphsize[graphnum[v] = cnt] ++;
	};
	bool* visited = new bool[n];
	// split into subgraphs
	for(int i=0; i<n; i++) visited[i] = false;
	for(int i=0; i<n; i++)
		if(!visited[i])
		{
    
    
			bfs(G, i, vis, visited);
			cnt ++;
		}
	delete[] visited;
}

int main(int argc, char** argv)
{
    
    
	// Input
	scanf("%d%d%d", &n, &m, &k);
	for(int i=0; i<m; i++)
	{
    
    
		int x, y;
		scanf("%d%d", &x, &y);
		friends[--x].push_back(--y);
		friends[y].push_back(x);
	}
	for(int i=0; i<k; i++)
	{
    
    
		int x, y;
		scanf("%d%d", &x, &y);
		block[--x].push_back(--y);
		block[y].push_back(x);
	}
	// Split firends into subgraphs
	splitgraph(friends);
	// Count bad (friendships & blockships in current subgraph & itself) for each vertex
	int* badcnt = new int[n];
	for(int i=0; i<n; i++) badcnt[i] = 1; // 1, not 0! (the vertex itself)
	for(int i=0; i<n; i++)
	{
    
    
		// Blockship 
		for(int it: block[i])
			if(it > i && graphnum[it] == graphnum[i])
				badcnt[i] ++, badcnt[it] ++;
		// Friendship
		for(int it: friends[i])
			if(it > i && graphnum[it] == graphnum[i])
				badcnt[i] ++, badcnt[it] ++;
	}
	for(int i=0; i<n; i++) printf("%d ", subgraphsize[graphnum[i]] - badcnt[i]);
	putchar('\n');
	delete[] badcnt;
	return 0;
}