HDU - 5876 :Sparse Graph (完全图的补图的最短路 -BFS&set)
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
InputThere 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.OutputFor 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
题意:给定一个无向图,求它的补图中S到每一点的最短路。
思路:我们BFS,长度从0,1,2...慢慢试探,由于是补图,显然试探的次数不会太多就可以弄完,所以我们可以暴力一点,一次BFS,我们取出队首u,其最短距离是dis[u],那么它没有被访问的点中(满足dis==-1),不与u相邻的点的最短距离是dis[u]+1,将其加入队首; 我们可以用set来表示未被访问的点。
由于S1.swap(S2)是两个set的指针交换,所以复杂度是O(1),比较快的。主要复杂度再S.clear那里,clear的复杂度是元素个数,由于是补图,所以可以假设放进去之后几个回合内就删完了,复杂度不会太高。
#include<bits/stdc++.h> #define rep(i,a,b) for(int i=a;i<=b;i++) using namespace std; const int maxn=200010; int Laxt[maxn],Next[maxn],To[maxn],dis[maxn],num[maxn]; int q[maxn],tot,cnt,S,N; void add(int u,int v){ Next[++cnt]=Laxt[u]; Laxt[u]=cnt; To[cnt]=v; } void BFS() { dis[S]=0; set<int>S1,S2; set<int>::iterator it; queue<int>q; q.push(S); rep(i,1,N) if(i!=S) S1.insert(i); while(!q.empty()){ int u=q.front(); q.pop(); for(int i=Laxt[u];i;i=Next[i]){ int v=To[i]; if(S1.find(v)==S1.end()) continue; S1.erase(v); S2.insert(v); } for(it=S1.begin();it!=S1.end();it++) dis[*it]=dis[u]+1,q.push(*it); S1.swap(S2); S2.clear(); } } int main() { int T,M,u,v; scanf("%d",&T); while(T--){ scanf("%d%d",&N,&M); cnt=0; rep(i,1,N) Laxt[i]=0,dis[i]=-1; rep(i,1,M){ scanf("%d%d",&u,&v); add(u,v); add(v,u); } scanf("%d",&S); BFS(); rep(i,1,N){ if(i!=S){ if(i!=N) printf("%d ",dis[i]); else printf("%d\n",dis[i]); } } } return 0; }