hdu 5876 Sparse Graph 无权图bfs求最短路
Posted xjhz
tags:
篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了hdu 5876 Sparse Graph 无权图bfs求最短路相关的知识,希望对你有一定的参考价值。
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 other vertices.
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 other vertices.
Input
There are multiple test cases. The first line of input is an integer T denoting the number of test cases. For each test case, the first line contains two integers N and M. The following M lines each contains two distinct integers u,v(1≤u,v≤N denoting an edge. And S is given on the last line.
Output
For each of T test cases, print a single line consisting of N space separated integers, denoting shortest distances of the remaining N 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
#include<bits/stdc++.h> using namespace std; #define ll long long #define pi (4*atan(1.0)) const int N=1e5+10,M=4e6+10,inf=1e9+10,mod=1e9+7; const ll INF=1e18+10; vector<int>v[N<<1]; queue<int>q; set<int>s; int flag[N<<1]; int ans[N<<1]; void init(int n) { s.clear(); for(int i=1;i<=n;i++) v[i].clear(),flag[i]=1,ans[i]=0; while(!q.empty())q.pop(); } int main() { int T; scanf("%d",&T); while(T--) { int n,m; scanf("%d%d",&n,&m); init(n); for(int i=0;i<m;i++) { int u,w; scanf("%d%d",&u,&w); v[u].push_back(w); v[w].push_back(u); } int st; scanf("%d",&st); q.push(st); for(int i=1;i<=n;i++)if(i!=st)s.insert(i); flag[st]=0,ans[st]=0; while(!q.empty()) { int u=q.front(); q.pop(); for(int i=0;i<v[u].size();i++) flag[v[u][i]]=0; for(set<int>::iterator itt,it=s.begin();it!=s.end();) { if(flag[*it]) { q.push(*it); ans[*it]=ans[u]+1; itt=it; it++; s.erase(itt); } else it++; } for(set<int>::iterator it=s.begin();it!=s.end();it++)flag[*it]=1; } int flag=0; for(int i=1;i<=n;i++) { if(i==st) continue; printf("%d%c",ans[i],(flag++>=(n-2)?‘\n‘:‘ ‘)); } } return 0; }
以上是关于hdu 5876 Sparse Graph 无权图bfs求最短路的主要内容,如果未能解决你的问题,请参考以下文章