HDU 5876 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 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

求最短路，由于是完全图上删边，显然最短路不会太长，所以暴力的扫描即可。
#include<set>
#include<map>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#define rep(i,j,k) for (int i = j; i <= k; i++)
#define per(i,j,k) for (int i = j; i >= k; i--)
#define loop(i,j,k) for (int i = j;i != -1; i = k[i])
#define lson x << 1, l, mid
#define rson x << 1 | 1, mid + 1, r
#define ff first
#define ss second
#define mp(i,j) make_pair(i,j)
#define pb push_back
#define pii pair<int,int>
#define in(x) scanf("%d", &x);
using namespace std;
typedef long long LL;
const int low(int x) { return x&-x; }
const double eps = 1e-8;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int N = 2e5 + 10;
int T, n, m, x ,y;
int ft
,nt
,u
,sz;
int dis
;

int main()
{
in(T);
while (T--)
{
sz=0;
scanf("%d%d",&n,&m);
rep(i,1,n) dis[i]=-1,ft[i]=-1;
rep(i,1,m)
{
scanf("%d%d",&x,&y);
u[sz]=y; nt[sz]=ft[x]; ft[x]=sz++;
u[sz]=x; nt[sz]=ft[y]; ft[y]=sz++;
}
scanf("%d",&x);
dis[x]=0;
int sum=n-1,t=0;
while (sum)
{
++t;
int q=0;
rep(i,1,n)
{
if (dis[i]!=-1) continue;
int res=0,ans=0;
loop(j,ft[i],nt)
{
if (dis[u[j]]!=-1&&dis[u[j]]<t) res++;
}
if (res+sum==n) continue;
else {dis[i]=t; q++;}
}
sum-=q;
if (!q) break;
}
int flag=0;
rep(i,1,n)
{
if (dis[i])
{
printf("%s%d",flag?" ":"",dis[i]);
flag=1;
}
}
putchar(10);
}
return 0;
}