2016大连网络赛 Sparse Graph
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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−1 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−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
分析:对于能够到达的点依次放入队列,暴力未放入的点,可行继续放入队列即可;
代码:
#include <iostream> #include <cstdio> #include <cstdlib> #include <cmath> #include <algorithm> #include <climits> #include <cstring> #include <string> #include <set> #include <map> #include <queue> #include <stack> #include <vector> #include <list> #define rep(i,m,n) for(i=m;i<=n;i++) #define rsp(it,s) for(set<int>::iterator it=s.begin();it!=s.end();it++) #define mod 1000000007 #define inf 0x3f3f3f3f #define vi vector<int> #define pb push_back #define mp make_pair #define fi first #define se second #define ll long long #define pi acos(-1.0) #define pii pair<int,int> #define Lson L, mid, rt<<1 #define Rson mid+1, R, rt<<1|1 const int maxn=2e5+10; const int dis[4][2]={{0,1},{-1,0},{0,-1},{1,0}}; using namespace std; ll gcd(ll p,ll q){return q==0?p:gcd(q,p%q);} ll qpow(ll p,ll q){ll f=1;while(q){if(q&1)f=f*p%mod;p=p*p%mod;q>>=1;}return f;} int n,m,k,t,dp[maxn]; set<int>a,b[maxn],c; queue<int>p; int main() { int i,j; scanf("%d",&t); while(t--) { bool flag=false; scanf("%d%d",&n,&m); memset(dp,-1,sizeof dp); rep(i,1,n)b[i].clear(),a.insert(i); while(m--) { int u,v; scanf("%d%d",&u,&v); b[u].insert(v),b[v].insert(u); } scanf("%d",&m); p.push(m);dp[m]=0;a.erase(m); while(!p.empty()) { c.clear(); int u=p.front();p.pop(); for(int x:a)if(b[x].find(u)==b[x].end())dp[x]=dp[u]+1,c.insert(x),p.push(x); for(int x:c)a.erase(x); } rep(i,1,n)if(i!=m) { if(flag)printf(" %d",dp[i]); else printf("%d",dp[i]),flag=true; } printf("\n"); } //system("pause"); return 0; }
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