HDU 5876：Sparse Graph（BFS）

Posted 2020-08-08 Shadowdsp

http://acm.hdu.edu.cn/showproblem.php?pid=5876

Sparse Graph

 

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

 

题意：给出一个图，和一个起点，求在该图的补图中从起点到其他N-1个点的最短距离。如果不连通输出-1.

思路：比赛的时候觉得这题N那么大不会做。现在回过头发现貌似不难。用set将未走过的点放置进去，并在对点的邻边进行扩展的时候，把能走到的邻点删除掉（即补图中可以走到的邻点保留）。

 1 #include <cstdio>
 2 #include <algorithm>
 3 #include <iostream>
 4 #include <cstring>
 5 #include <string>
 6 #include <cmath>
 7 #include <queue>
 8 #include <vector>
 9 #include <set>
10 using namespace std;
11 #define N 200010
12 #define INF 0x3f3f3f3f
13 
14 struct node
15 {
16     int v, nxt;
17 }edge[N];
18 int head[N];
19 int d[N], tot;
20 
21 void add(int u, int v)
22 {
23     edge[tot].v = v; edge[tot].nxt = head[u]; head[u] = tot++;
24     edge[tot].v = u; edge[tot].nxt = head[v]; head[v] = tot++;
25 }
26 
27 void bfs(int st, int n)
28 {
29     queue<int> que;
30     d[st] = 0;
31     que.push(st);
32     set<int> s1, s2;
33     for(int i = 1; i <= n; i++) {
34         if(i != st) s1.insert(i);
35     }
36     while(!que.empty()) {
37         int u = que.front(); que.pop();
38         for(int k = head[u]; ~k; k = edge[k].nxt) {
39             int v = edge[k].v;
40             if(!s1.count(v)) continue;
41             s1.erase(v); //补图中当前能走到的并且还未更新过的
42             s2.insert(v); //补图中走不到的还要扩展的
43         }
44         for(set<int>:: iterator it = s1.begin(); it != s1.end(); it++) {
45             d[*it] = d[u] + 1;
46             que.push(*it);
47         }
48         s1.swap(s2); //还没更新过的
49         s2.clear();
50     }
51 }
52 
53 int main()
54 {
55     int t;
56     scanf("%d", &t);
57     while(t--) {
58         memset(head, -1, sizeof(head));
59         memset(d, INF, sizeof(INF));
60         int n, m;
61         scanf("%d%d", &n, &m);
62         for(int i = 0; i < m; i++) {
63             int u, v;
64             scanf("%d%d", &u, &v);
65             add(u, v);
66         }
67         int st;
68         scanf("%d", &st);
69         bfs(st, n);
70         bool f = 0;
71         for(int i = 1; i <= n; i++) {
72             if(i != st) {
73                 if(f) printf(" ");
74                 f = 1;
75                 if(d[i] == INF) printf("-1");
76                 else printf("%d", d[i]);
77             }
78         }
79         puts("");
80     }
81 
82     return 0;
83 }