HDU 5876 Sparse Graph

Sparse GraphTime Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1489    Accepted Submission(s): 521Problem DescriptionIn 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.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 Input12 01Sample Output1

水题啊…..英语不好 网赛的时候瞄了几眼 就没看下去了…..
用链表记录还没到过的点 到一个删一个
set<pair<int,int> >记录原图中有的边(不可通过的边)
BFS跑一遍 其实最多才进行n+m次松弛操作

#include<iostream>#include<stdlib.h>#include<stdio.h>#include<string>#include<vector>#include<deque>#include<queue>#include<algorithm>#include<set>#include<map>#include<stack>#include<time.h>#include<math.h>#include<list>#include<cstring>#include<fstream>#include<bitset>//#include<memory.h>using namespace std;#define ll long long#define ull unsigned long long#define pii pair<int,int>#define INF 1000000007const int N=200000+5;set<pii>exist;int dis[N];list<int>ls;void bfs(int s,int n){    fill(dis,dis+n+1,-1);    ls.clear();    for(int i=1;i<=n;++i)        if(i!=s)            ls.push_back(i);    deque<int>de;    de.push_back(s);    dis[s]=0;    while(!de.empty()){        int fr=de.front();        de.pop_front();        if(ls.empty())            break;        for(list<int>::iterator it=ls.begin();it!=ls.end();){            if(exist.find({*it,fr})!=exist.end())                ++it;            else{                de.push_back(*it);                dis[*it]=dis[fr]+1;                list<int>::iterator tmp=it;                ++it;                ls.erase(tmp);            }        }    }}int main(){    //freopen("/home/lu/文档/r.txt","r",stdin);    //freopen("/home/lu/文档/w.txt","w",stdout);    int t,n,m,u,v,s;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&m);        exist.clear();        while(m--){            scanf("%d%d",&u,&v);            exist.insert({u,v});            exist.insert({v,u});        }        scanf("%d",&s);        bfs(s,n);        bool flag=false;        for(int i=1;i<=n;++i){            if(i!=s){                if(flag==false)                    flag=true;                else                    printf(" ");                printf("%d",dis[i]);            }        }        putchar('\n');    }    return 0;}