HDU 5876 Sparse Graph
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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.
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
解析:补图上的BFS。
#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, s2; set<int>::iterator it; for(int i = 1; i <= n; ++i){ 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); s2.insert(v); } for(it = s1.begin(); it != s1.end(); ++it){ q.push(*it); dis[*it] = dis[u]+1; } s1.swap(s2); s2.clear(); } } 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("\n"); } int main() { int t; scanf("%d", &t); while(t--){ scanf("%d%d", &n, &m); solve(); } return 0; }
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