HDU 5876 补图 单源 最短路

Posted 2020-10-28 灬从此以后灬

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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 }

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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 }

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