HDU 5876 Sparse Graph

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本文探讨了图论中图补的概念，并提出了一种计算给定顶点到图补中其它所有顶点最短距离的方法。通过遍历和特定算法优化，在大规模图中实现了高效计算。

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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[N],nt[N],u[N],sz;
int dis[N];

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;
}