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Sparse Graph
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2590 Accepted Submission(s): 902
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 求源点 s 在补图中到其它点的最短路长
解析 因为 给的边比较少 点比较多 补图就是一个非常稠密的图 我们不能建补图 迪杰斯特拉跑一边 会超时的 我们注意到 边权都为1 我们可以直接BFS一层一层求最短路
每个点都入队一次 然后取队顶在未被访问过的点中找在原图中不与其相邻的点 加入队列 。因为在原图中不与点s相连的点 在补图中最短路都为1 待访问的不超过m个
所以bfs是可行的 但是我们在找 与 当前队顶其不相邻的点 用数组标记 花费时间很大（每次遍历n会超时）我们用set存一下 就好了 只留下待访问的点 时间复杂度就降下来了
set AC代码
1 #include <iostream>
2 #include <stdio.h>
3 #include <algorithm>
4 #include <string.h>
5 #include <stdlib.h>
6 #include <string>
7 #include <queue>
8 #include <map>
9 #include <set>
10 #include <vector>
11 using namespace std;
12 const int maxn=2e5+10;
13 int n,m,s;
14 vector<int> g[maxn];
15 int ans[maxn],vis[maxn];
16 set<int> a,b;
17 void bfs(int x)
18 {
19 for(int i=1;i<=n;i++)
20 if(s!=i)
21 a.insert(i);
22 queue<int> q;
23 q.push(x);
24 while(!q.empty())
25 {
26 int u=q.front();
27 q.pop();
28 for(int i=0;i<g[u].size();i++)
29 {
30 int v=g[u][i];
31 if(a.count(v))
32 {
33 b.insert(v);
34 a.erase(v);
35 }
36 }
37 set<int>::iterator it;
38 for(it=a.begin();it!=a.end();it++)
39 {
40 q.push(*it);
41 ans[*it]=ans[u]+1;
42 }
43 a.swap(b);
44 b.clear();
45 }
46 }
47 int main()
48 {
49 int t;
50 cin>>t;
51 while(t--)
52 {
53 cin>>n>>m;
54 a.clear(),b.clear();
55 for(int i=1;i<=n;i++)
56 {
57 g[i].clear();
58 }
59 memset(ans,-1,sizeof(ans));
60 for(int i=0;i<m;i++)
61 {
62 int u,v;
63 cin>>u>>v;
64 g[u].push_back(v);
65 g[v].push_back(u);
66 }
67 cin>>s;
68 ans[s]=0;
69 // set<int>::iterator j;
70 // for(j=a.begin();j!=a.end();j++)
71 // cout<<*j<<endl;
72 bfs(s);
73 if(s!=n)
74 for(int i=1;i<=n;i++)
75 {
76 if(i!=s)
77 {
78 if(i!=n)
79 cout<<ans[i]<<" ";
80 else
81 cout<<ans[n]<<endl;
82 }
83 }
84 else
85 for(int i=1;i<=n-1;i++)
86 {
87 if(i==n-1)
88 cout<<ans[n-1]<<endl;
89 else
90 cout<<ans[i]<<" ";
91 }
92 }
93 return 0;
94 }
数组超时代码
1 #include <iostream>
2 #include <stdio.h>
3 #include <algorithm>
4 #include <string.h>
5 #include <stdlib.h>
6 #include <string>
7 #include <queue>
8 #include <map>
9 #include <vector>
10 using namespace std;
11 const int maxn=2e5+10;
12 int n,m,s;
13 vector<int> g[maxn];
14 int ans[maxn],vis[maxn],a[maxn];
15 void bfs(int x)
16 {
17 vis[x]=1;
18 queue<int> q;
19 q.push(x);
20 int cnt=1;
21 while(!q.empty())
22 {
23 int u=q.front();
24 q.pop();
25 for(int i=0;i<g[u].size();i++)
26 {
27 int v=g[u][i];
28 if(cnt%2==1)
29 a[v]=0;
30 else
31 a[v]=1;
32 }
33 for(int i=1;i<=n;i++)
34 {
35 if(cnt%2==1&&a[i]==1&&!vis[i])
36 {
37 ans[i]=ans[u]+1;
38 q.push(i);
39 vis[i]=1;
40 }
41 if(cnt%2==0&&a[i]==0&&!vis[i])
42 {
43 ans[i]=ans[u]+1;
44 q.push(i);
45 vis[i]=1;
46 }
47 }
48 cnt++;
49 }
50 }
51 int main()
52 {
53 int t;
54 cin>>t;
55 while(t--)
56 {
57 cin>>n>>m;
58 for(int i=1;i<=n;i++)
59 g[i].clear(),a[i]=1;
60 memset(ans,-1,sizeof(ans));
61 memset(vis,0,sizeof(vis));
62 for(int i=0;i<m;i++)
63 {
64 int u,v;
65 cin>>u>>v;
66 g[u].push_back(v);
67 g[v].push_back(u);
68 }
69 cin>>s;ans[s]=0;
70 bfs(s);
71 for(int i=1;i<=n;i++)
72 {
73 if(i!=s)
74 cout<<ans[i]<<" ";
75 }
76 cout<<endl;
77 }
78 return 0;
79 }
原文地址：https://www.cnblogs.com/stranger-/p/8666410.html