2016大连网络赛 Sparse Graph

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)

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

 

分析：对于能够到达的点依次放入队列，暴力未放入的点，可行继续放入队列即可；

代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <climits>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <vector>
#include <list>
#define rep(i,m,n) for(i=m;i<=n;i++)
#define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++)
#define mod 1000000007
#define inf 0x3f3f3f3f
#define vi vector<int>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define ll long long
#define pi acos(-1.0)
#define pii pair<int,int>
#define Lson L, mid, rt<<1
#define Rson mid+1, R, rt<<1|1
const int maxn=2e5+10;
const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}};
using namespace std;
ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);}
ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;}
int n,m,k,t,dp[maxn];
set<int>a,b[maxn],c;
queue<int>p;
int main()
{
    int i,j;
    scanf("%d",&t);
    while(t--)
    {
        bool flag=false;
        scanf("%d%d",&n,&m);
        memset(dp,-1,sizeof dp);
        rep(i,1,n)b[i].clear(),a.insert(i);
        while(m--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            b[u].insert(v),b[v].insert(u);
        }
        scanf("%d",&m);
        p.push(m);dp[m]=0;a.erase(m);
        while(!p.empty())
        {
            c.clear();
            int u=p.front();p.pop();
            for(int x:a)if(b[x].find(u)==b[x].end())dp[x]=dp[u]+1,c.insert(x),p.push(x);
            for(int x:c)a.erase(x);
        }
        rep(i,1,n)if(i!=m)
        {
            if(flag)printf(" %d",dp[i]);
            else printf("%d",dp[i]),flag=true;
        }
        printf("
");
    }
    //system("pause");
    return 0;
}