HDU 5876 Sparse Graph

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 1520    Accepted Submission(s): 537

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 notadjacent 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

 

Source

2016 ACM/ICPC Asia Regional Dalian Online

 

 

 

解析：补图上的最短路。维护一个set，里面保存尚未访问的结点。BFS的扩展结点u的时候，把与u直接相连的结点排除，得到的set在补图中都与u有一条权为1的边，访问完之后把这些点从set中删除，开始下一次扩展。即在扩展结点u的时候，只扩展与u没有边相连的结点，有边相邻的结点等下一次再扩展。

 

 

 

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

const int INF = 0x3f3f3f3f;
const int MAXN = 2e5+5;
vector<int> e[MAXN];
int dis[MAXN];
int n, m, s;

void bfs()
{
    set<int> s1;    //保存还未被访问过的结点
    set<int> s2;    //保存与节点u相连的结点
    set<int>::iterator it;
    for(int i = 1; i <= n; ++i){    //s1初始化为除结点s外的所有结点
        if(i != s)
            s1.insert(i);
    }
    queue<int> q;
    q.push(s);
    dis[s] = 0;
    while(!q.empty()){
        int u = q.front();
        q.pop();
        int len = e[u].size();
        for(int i = 0; i < len; ++i){
            int v = e[u][i];
            if(s1.find(v) == s1.end())
                continue;
            s1.erase(v);    //把与结点u之间有边的点从s1排除
            s2.insert(v);   //并加入到s2
        }
        for(it = s1.begin(); it != s1.end(); ++it){
            q.push(*it);
            dis[*it] = dis[u]+1;
        }
        s1.swap(s2);    //s1中的结点已经全部访问，s2中的结点尚未访问，二者交换
        s2.clear(); //清空
    }
}

void solve()
{
    memset(dis, INF, sizeof(dis));
    for(int i = 1; i <= n; ++i)
        e[i].clear();
    int u, v;
    while(m--){
        scanf("%d%d", &u, &v);
        e[u].push_back(v);
        e[v].push_back(u);
    }
    scanf("%d", &s);
    bfs();
    for(int i = 1; i <= n; ++i){
        if(i == s)
            continue;
        if(dis[i] >= INF)
            dis[i] = -1;
        printf("%d", dis[i]);
        if(i != n)
            printf(" ");
    }
    printf("
");
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d%d", &n, &m);
        solve();
    }
    return 0;
}