Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2830    Accepted Submission(s): 972

 

Problem Description

In graph theory, the complement of a graph G is a graph H on the same vertices such that two distinct vertices of H are adjacent if and only if they are not adjacent in G .

Now you are given an undirected graph G of N nodes and M bidirectional edges of unit length. Consider the complement of G , i.e., H . For a given vertex S on H , you are required to compute the shortest distances from S to all N−1 other vertices.

 

 

Input

There are multiple test cases. The first line of input is an integer T(1≤T<35) denoting the number of test cases. For each test case, the first line contains two integers N(2≤N≤200000) and M(0≤M≤20000) . The following M lines each contains two distinct integers u,v(1≤u,v≤N) denoting an edge. And S (1≤S≤N) is given on the last line.

 

 

Output

For each of T test cases, print a single line consisting of N−1 space separated integers, denoting shortest distances of the remaining N−1 vertices from S (if a vertex cannot be reached from S, output ``-1" (without quotes) instead) in ascending order of vertex number.

 

 

Sample Input


 

1 2 0 1

 

 

Sample Output


 

1

 

       在我们知道一个图的时候，再求补图的连接情况暴力是非常耗时的（这不废话），为了进行优化，我们通过set来进行补图时的路的延伸。

        这样的话，我们设置两个set:s1和s2, s1代表在这一轮补图中的延伸中可以被更新距离的点的集合，s2代表在到现在的所有的延伸中仍然无法更新距离值得点得集合，这样通过set得运用可以极大得避免一些不必要的查找。

#include<bits/stdc++.h>
using namespace std;
int m, n;
const int maxn = 200010;
const int inf = 0x3f3f3f3f;
struct *** {
	int u, v, ne;
}ed[maxn];
int head[maxn], cnt = 0, d[maxn];
void init() {
	memset(head, -1, sizeof(head));
	memset(d, inf, sizeof(d));
	cnt = 0;
}
void add(int u, int v) {
	ed[cnt].u = u; ed[cnt].v = v;
	ed[cnt].ne = head[u]; head[u] = cnt++;
}
void bfs(int st) {
	queue<int>q;
	set<int>s1, s2;
	d[st] = 0;
	q.push(st);
	for (int s = 1; s <= n; s++)
		if (s != st)s1.insert(s);
	while (!q.empty()){
		int u = q.front(); q.pop();
		for (int s = head[u]; ~s; s = ed[s].ne) {
			int v = ed[s].v;
			if (!s1.count(v))continue;
			s1.erase(v);
			s2.insert(v);
		}
		for (set<int>::iterator now = s1.begin(); now != s1.end(); now++){
			d[*now] = d[u] + 1;
			q.push(*now);
		}
		s1.swap(s2);
		s2.clear();
	}
}
int main()
{
	int te;
	scanf("%d", &te);
	while (te--) {
		scanf("%d %d", &n, &m);
		init();
		while (m--) {
			int a, b;
			scanf("%d%d", &a, &b);
			add(a, b); add(b, a);
		}
		int st;
		scanf("%d", &st);
		bfs(st); int t = 0;
		for (int s = 1; s <= n; s++) {
			if (d[s] == inf)d[s] = -1;
			if (s == st)continue;
			if (!t) {
				printf("%d", d[s] );
				t = 1;
			}
			else {
				printf(" %d", d[s]);
			}
		}
		printf("\n");
	}
	return 0;
}