HDU 5876 Sparse Graph(2016 ACM/ICPC Asia Regional Dalian Online)
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题目大意
给出一个图,在这个图的补图中给定一个点,求这个点到其他所有点的距离,如果无法到达则输出-1.
题目分析
因为这个图的点数比较多,因此遍历所有边很明显是不可能的,铁定超时,因此只能想一些办法来处理,我们可以将所有遍历过的点塞入链表中,如果每一次有bfs求出一些节点的值之后,那么则将这些节点从链表中删除,这样每次遍历的点将会少很多,这样就可以在O(n+m)的时间内得到结果。
#include <set>#include <list>#include <queue>#include <vector>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 2e5+5;typedef pair<int,int> PII;int dist[maxn];list <int> LIST;set <PII> SET;vector <list <int>::iterator > VEC;vector <int> vec;void bfs(int S){ queue <int> que; que.push(S); while(!que.empty()){ int u = que.front(); que.pop(); VEC.clear(); for(list<int>::iterator it = LIST.begin(); it != LIST.end(); it++){ int v = *it; if(SET.count(make_pair(u, v)) == 0){ dist[v] = dist[u] + 1; que.push(v); VEC.push_back(it); } } for(int i = 0; i < VEC.size(); i++) LIST.erase(VEC[i]); }}int main(){ int T,N,M,S; scanf("%d", &T); while(T--){ memset(dist, 0, sizeof(dist)); scanf("%d%d", &N, &M); int from, to; SET.clear(); while(M--){ scanf("%d%d", &from, &to); SET.insert(make_pair(from, to)); SET.insert(make_pair(to, from)); } scanf("%d", &S); for(int i = 1; i <= N; i++) if(i != S) LIST.push_back(i); bfs(S); vec.clear(); for(int i = 1; i <= N; i++) if(i != S) vec.push_back(dist[i]); for(int i = 0; i < vec.size(); i++) if(i != vec.size()-1){ if(vec[i]) printf("%d ", vec[i]); else printf("-1 "); } else{ if(vec[i]) printf("%d\n", vec[i]); else printf("-1\n"); } } return 0;}
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