hdu5876 Sparse Graph -补图的最短路-bfs

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Sparse Graph

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 integerT(1≤T<35) denoting the number of test cases. For each test case, the first line contains two integersN(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

Source

2016 ACM/ICPC Asia Regional Dalian Online

题目大意

给出一个图，求这个图的补图的最短路，每条路的长度都为1；

解题思路

一个图的补图就是，把这个图补成完全图的图，n阶完全图是每个点的度数都为n-1

能确定的是与起点不连通的点最短距离一定是1，与起点连通的点最短距离不确定，只能用与起点不连通的点去更新，
假如起点s与点a连通与点b不连通且点a与点b不连通，那么a点的最小值就可以取决于点b，
也就是说与起点连通的点a的最短距离取决于与起点不连通并且与点a不连通的所有点中距离最小的一个+1
所以用到bfs 先1步找到与起点不连通的点，再2步找到与与起点不连通的点不连通的点（好绕），依次类推就能找到每个点的最小距离
还有输出格式的问题，注意一个特例就是起点是最后一个点

代码

#include <cstdio>#include <iostream>#include <cstring>#include <stack>#include <set>#include <queue>#include <algorithm>using namespace std;const int maxn = 200010;int n,s;int dist[maxn];vector <int> g[maxn];queue <int > q;void f(){    set<int> a,b;    q.push(s);    for (int i = 1;i <= n;i ++)    {       if (i != s)            a.insert(i);    }    while(!q.empty())    {        int now=q.front();        q.pop();        for(int i=0;i<g[now].size();++i)        {            int v=g[now][i];            if(!a.count(v)) continue;            a.erase(v);            b.insert(v);        }        for(set<int>::iterator it = a.begin();it!=a.end();++it)        {            dist[*it]=dist[now]+1;//以起点为例，与起点不相邻的点距离一定是0+1，            q.push(*it);//每次放入队列的都是与之不相邻的点。这样保证这点是最小距离        }        //把和当前点相邻的点放入A集合，使得与当前点不相邻的点去更新它们        a.swap(b);        b.clear();    }    for(int i=1;i<=n;++i)    {        if(s==i) continue;        printf("%d",dist[i]);        if(s==n && i==n-1) continue;        if(i!=n) printf(" ");    }    printf("\n");}int main(){    int t;    scanf("%d",&t);    while(t--){        int m;        scanf("%d%d",&n,&m);        for (int i = 1;i <= n;i ++)            g[i].clear();        memset(dist,0,sizeof(dist));        for (int i = 1;i <= m;i ++)        {            int u,v;            scanf("%d%d",&u,&v);            g[u].push_back(v);            g[v].push_back(u);        }        scanf("%d",&s);        f();    }    return 0;}

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