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+10;
int n,m,s;
vector<int> g[maxn];
int ans[maxn],vis[maxn];
set<int> a,b;
void bfs(int x)
{
    for(int i=1;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=0;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]+1;
        }
        a.swap(b);
        b.clear();
    }
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        a.clear(),b.clear();
        for(int i=1;i<=n;i++)
        {
            g[i].clear();
        }
        memset(ans,-1,sizeof(ans));
        for(int i=0;i<m;i++)
        {
            int u,v;
            cin>>u>>v;
            g[u].push_back(v);
            g[v].push_back(u);
        }
        cin>>s;
        ans[s]=0;
 //       set<int>::iterator j;
//        for(j=a.begin();j!=a.end();j++)
//            cout<<*j<<endl;
        bfs(s);
        if(s!=n)
            for(int i=1;i<=n;i++)
            {
                if(i!=s)
                {
                    if(i!=n)
                        cout<<ans[i]<<" ";
                    else
                        cout<<ans[n]<<endl;
                }
            }
        else
            for(int i=1;i<=n-1;i++)
            {
                if(i==n-1)
                    cout<<ans[n-1]<<endl;
                else
                    cout<<ans[i]<<" ";
            }
    }
    return 0;
}

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+10;
int n,m,s;
vector<int> g[maxn];
int ans[maxn],vis[maxn],a[maxn];
void bfs(int x)
{
    vis[x]=1;
    queue<int> q;
    q.push(x);
    int cnt=1;
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        for(int i=0;i<g[u].size();i++)
        {
            int v=g[u][i];
            if(cnt%2==1)
                a[v]=0;
            else
                a[v]=1;
        }
        for(int i=1;i<=n;i++)
        {
            if(cnt%2==1&&a[i]==1&&!vis[i])
            {
                ans[i]=ans[u]+1;
                q.push(i);
                vis[i]=1;
            }
            if(cnt%2==0&&a[i]==0&&!vis[i])
            {
                ans[i]=ans[u]+1;
                q.push(i);
                vis[i]=1;
            }
        }
        cnt++;
    }
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        cin>>n>>m;
        for(int i=1;i<=n;i++)
            g[i].clear(),a[i]=1;
        memset(ans,-1,sizeof(ans));
        memset(vis,0,sizeof(vis));
        for(int i=0;i<m;i++)
        {
            int u,v;
            cin>>u>>v;
            g[u].push_back(v);
            g[v].push_back(u);
        }
        cin>>s;ans[s]=0;
        bfs(s);
        for(int i=1;i<=n;i++)
        {
            if(i!=s)
                cout<<ans[i]<<" ";
        }
        cout<<endl;
    }
    return 0;
}