hdu5876 Sparse Graph（最短路）

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思路：一个很经典的题目，求补图的最短路，方法是建原图，然后用两个set维护还没 BFS 过的点。当要扩展点 u 的时候，遍历一次还没访问过的点 v，如果 uv 没边，那么将 v 入队。否则将 v 留在未扩展点中。

#include<bits/stdc++.h>using namespace std;const int maxn = 200000+7;vector<int>e[maxn];int n,m,s,dis[maxn];set<int>v1,v2;set<int>::iterator it;int main(){    int T;scanf("%d",&T);while(T--){        scanf("%d%d",&n,&m);for(int i = 0;i<=n;i++)e[i].clear();memset(dis,-1,sizeof(dis));v1.clear();v2.clear();for(int i = 1;i<=m;i++){int u,v;scanf("%d%d",&u,&v);e[u].push_back(v);e[v].push_back(u);}for(int i = 1;i<=n;i++)v1.insert(i);scanf("%d",&s);queue<int>q;q.push(s);dis[s]=0;v1.erase(s);        while(!q.empty()){             int u = q.front();q.pop(); for(int i = 0;i<e[u].size();i++) { int v = e[u][i]; if(!v1.count(v)) continue; v1.erase(v); v2.insert(v); } for(it = v1.begin();it!=v1.end();it++) { q.push(*it); dis[*it]=dis[u]+1; } v1.swap(v2); v2.clear();}for(int i = 1;i<=n;i++){if(i==s)continue;printf("%d%c",dis[i],i==n?'\n':' ');}}}

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

Source

2016 ACM/ICPC Asia Regional Dalian Online

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