HDU 5876 Sparse Graph BFS 最短路

Posted 2020-08-09 Meek

Sparse Graph

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.

 

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

 

 

题意：

　　

　　n 个点的无向完全图中删除 m 条边，问点 s 到其他点的最短路长度。

 

题解：

　　

　　

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
#include<queue>
#include<set>
using namespace std;

#pragma comment(linker, "/STACK:102400000,102400000")
#define ls i<<1
#define rs ls | 1
#define mid ((ll+rr)>>1)
#define pii pair<int,int>
#define MP make_pair

typedef long long LL;
const long long INF = 1e18;
const double Pi = acos(-1.0);
const int N = 2e5+10, M = 1e2+11, mod = 1e9+7, inf = 2e9;

 int T,n,m,vis[N],head[N],t,d[N],ans[N];
 struct ss{int to,next;}e[N * 2];
 void add(int u,int v) {e[t].to=v;e[t].next=head[u];head[u]=t++;}

int main() {
        scanf("%d",&T);
        while(T--) {
            scanf("%d%d",&n,&m);
            t =1;memset(head,0,sizeof(head));
            for(int i = 1; i <= m; ++i) {
                int a,b;
                scanf("%d%d",&a,&b);
                add(a,b);add(b,a);
            }
            memset(vis,0,sizeof(vis));
            memset(d,-1,sizeof(d));
            int S;
            scanf("%d",&S);
            queue<int > q;
            q.push(S);vis[S] = 1;
            d[S] = 0;
            set<int > s;
            for(int i = 1; i <= n; ++i) if(i != S) s.insert(i),vis[i] = 1;
            while(!q.empty()) {
                int k = q.front();
                q.pop();
                for(int i = head[k]; i; i =e[i].next) {
                    int to = e[i].to;
                    if(s.count(to)) {
                        vis[to] = 0;
                    }
                }
                for(set<int > ::iterator itt,it = s.begin();it != s.end(); ) {
                    if(vis[*it])
                    {
                        d[*it] = d[k] + 1;
                        q.push(*it);
                        itt = it;
                        itt++;
                        s.erase(it);
                        it = itt;
                    } else {
                        it++;
                    }
                }
                for(set<int > ::iterator itt,it = s.begin();it != s.end(); it++) vis[*it] = 1;
            }
             int cnt = 0;
            for(int i = 1; i <= n; ++i) {
                    if(i != S)ans[++cnt] = d[i];
            }
            for(int i = 1; i < cnt; ++i) printf("%d ",ans[i]);
            printf("%d\n",ans[cnt]);
        }
        return 0;
}