Sparse Graph hdu-5876（求补图的最短路）

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

    12 01 
  

Sample Output

1

Source

2016 ACM/ICPC Asia Regional Dalian Online

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<set>
#include<queue>
using namespace std;

const int maxm=2e4+100;
const int maxn=2e5+10;
const int inf=0x3f3f3f3f;


/*
思路真的是很强的，在我们的范围之内，但是要是想突破我们的固有思维
一看见最短路，脑子里就是各种模版，那真的是以前，交给我们模版，
到后来我们就没有模板了，怎样可以怎样来，大胆的猜想加上我们的验证
这就是我们要达到的高度，而且是必须达到的高度
*/


struct AE{
    int cnt;
    int head[maxn];

    struct Edge{
        int v,nt;
    }edge[maxm*2];

    void init(){
        this->cnt=0;
        memset(head,-1,sizeof(head));
    }

    void add_edge(int u,int v){
        edge[cnt].nt=head[u];
        edge[cnt].v=v;
        head[u]=cnt++;
    }
}ae;

int dis[maxn];

void bfs(int st,int n){
    queue<int> q;
    set<int> has_edge,no_edge;
    for(int i=1;i<=n;i++){
        dis[i]=inf;
        if(i==st)continue;
        no_edge.insert(i);
    }
    dis[st]=0;
    q.push(st);
    while(!q.empty()){
        int u=q.front();
        q.pop();
        for(int i=ae.head[u];i!=-1;i=ae.edge[i].nt){
            int v=ae.edge[i].v;
            if(!no_edge.count(v))continue;
            no_edge.erase(v);
            has_edge.insert(v);
        }
        for(set<int>::iterator it=no_edge.begin();it!=no_edge.end();it++){
            dis[*it]=dis[u]+1;
            q.push(*it);
        }
        no_edge=has_edge;
        has_edge.clear();
    }
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        ae.init();
        int n,m;
        scanf("%d %d",&n,&m);
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d %d",&u,&v);
            ae.add_edge(u,v);
            ae.add_edge(v,u);
        }
        int st;
        scanf("%d",&st);
        bfs(st,n);
        int flag=0;
        for(int i=1;i<=n;i++){
            if(i==st)continue;
            if(!flag)
                 printf("%d",dis[i]==inf?-1:dis[i]);
            else
                printf(" %d",dis[i]==inf?-1:dis[i]);
            flag++;
        }
        printf("\n");
    }
    return 0;
}

Sparse Graph hdu-5876（求补图的最短路）

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