hdu 5876 Sparse Graph（补图最短路）

Sparse Graph

                                                                       Time Limit: 4000/2000
MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)


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 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. 

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

2016 ACM/ICPC Asia Regional Dalian Online

题意：

题目给出一个由n个点，m条边组成的图，问如果在n 个点的无向完全图中删除这m 条边，点 s 到其他点的最短路长度是多少。

思路：

如果原图中u与s不相连，那么s->u的距离就是s+1，然后将u放入队列。

每次都找与当前节点不相连的点放入队列，然后求距离，这样求出来的距离肯定是最短的。

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

const int N=2e5+10;

set<int>G
, S[2];
set<int>:: iterator it;
queue<int>q;
int ans
;

void solve(int n, int s){
memset(ans, -1, sizeof(ans));
while(!q.empty()) q.pop();
ans[s]=0;
q.push(s);
int now=0;
for(int i=1; i<=n; i++){
if(i==s) continue;
S[now].insert(i);
}
while(!q.empty()){
s = q.front();q.pop();
S[!now].clear();
for(it=S[now].begin(); it!=S[now].end(); it++){
if(G[s].find(*it)==G[s].end()){
ans[*it] = ans[s]+1;
q.push(*it);
}
else{
S[!now].insert(*it);
}
}
now=!now;
}
}

int main(){
int t;
scanf("%d", &t);
while(t--){
int n, m, u, v, s;
scanf("%d%d", &n, &m);
for(int i=1; i<=n; i++) G[i].clear();
for(int i=0; i<m; i++){
scanf("%d%d", &u, &v);
G[u].insert(v);
G[v].insert(u);
}
scanf("%d", &s);
solve(n, s);
int len=0;
for(int i=1; i<=n; i++){
if(i==s) continue;
ans[len++] = ans[i];
}
for(int i=0; i<len; i++){
if(i==len-1) printf("%d\n", ans[i]);
else printf("%d ", ans[i]);
}
}
return 0;
}