HDU 5876 补图 单源 最短路

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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

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <string.h>
 #include <stdlib.h>
 #include <string>
 #include <queue>
 #include <map>
 #include <set>
 #include <vector>
 using namespace std;
 const int maxn=2e5+;
 int n,m,s;
 vector<int> g[maxn];
 int ans[maxn],vis[maxn];
 set<int> a,b;
 void bfs(int x)
 {
     for(int i=;i<=n;i++)
         if(s!=i)
             a.insert(i);
     queue<int> q;
     q.push(x);
     while(!q.empty())
     {
         int u=q.front();
         q.pop();
         for(int i=;i<g[u].size();i++)
         {
             int v=g[u][i];
             if(a.count(v))
             {
                 b.insert(v);
                 a.erase(v);
             }
         }
         set<int>::iterator it;
         for(it=a.begin();it!=a.end();it++)
         {
             q.push(*it);
             ans[*it]=ans[u]+;
         }
         a.swap(b);
         b.clear();
     }
 }
 int main()
 {
     int t;
     cin>>t;
     while(t--)
     {
         cin>>n>>m;
         a.clear(),b.clear();
         for(int i=;i<=n;i++)
         {
             g[i].clear();
         }
         memset(ans,-,sizeof(ans));
         for(int i=;i<m;i++)
         {
             int u,v;
             cin>>u>>v;
             g[u].push_back(v);
             g[v].push_back(u);
         }
         cin>>s;
         ans[s]=;
  //       set<int>::iterator j;
 //        for(j=a.begin();j!=a.end();j++)
 //            cout<<*j<<endl;
         bfs(s);
         if(s!=n)
             for(int i=;i<=n;i++)
             {
                 if(i!=s)
                 {
                     if(i!=n)
                         cout<<ans[i]<<" ";
                     else
                         cout<<ans[n]<<endl;
                 }
             }
         else
             for(int i=;i<=n-;i++)
             {
                 if(i==n-)
                     cout<<ans[n-]<<endl;
                 else
                     cout<<ans[i]<<" ";
             }
     }
     return ;
 }

数组超时代码

 #include <iostream>
 #include <stdio.h>
 #include <algorithm>
 #include <string.h>
 #include <stdlib.h>
 #include <string>
 #include <queue>
 #include <map>
 #include <vector>
 using namespace std;
 const int maxn=2e5+;
 int n,m,s;
 vector<int> g[maxn];
 int ans[maxn],vis[maxn],a[maxn];
 void bfs(int x)
 {
     vis[x]=;
     queue<int> q;
     q.push(x);
     int cnt=;
     while(!q.empty())
     {
         int u=q.front();
         q.pop();
         for(int i=;i<g[u].size();i++)
         {
             int v=g[u][i];
             if(cnt%==)
                 a[v]=;
             else
                 a[v]=;
         }
         for(int i=;i<=n;i++)
         {
             if(cnt%==&&a[i]==&&!vis[i])
             {
                 ans[i]=ans[u]+;
                 q.push(i);
                 vis[i]=;
             }
             if(cnt%==&&a[i]==&&!vis[i])
             {
                 ans[i]=ans[u]+;
                 q.push(i);
                 vis[i]=;
             }
         }
         cnt++;
     }
 }
 int main()
 {
     int t;
     cin>>t;
     while(t--)
     {
         cin>>n>>m;
         for(int i=;i<=n;i++)
             g[i].clear(),a[i]=;
         memset(ans,-,sizeof(ans));
         memset(vis,,sizeof(vis));
         for(int i=;i<m;i++)
         {
             int u,v;
             cin>>u>>v;
             g[u].push_back(v);
             g[v].push_back(u);
         }
         cin>>s;ans[s]=;
         bfs(s);
         for(int i=;i<=n;i++)
         {
             if(i!=s)
                 cout<<ans[i]<<" ";
         }
         cout<<endl;
     }
     return ;
 }