Hdu 5876 SparseGraph 2016ACM ICPC 大连网络赛

Sparse Graph

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)
Total Submission(s): 2612    Accepted Submission(s): 911

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

 

题意：给定一个N个顶点，M条边的无向图G，求其补图H中给定的源点S到其他N-1个点的最短距离，每条边的长度为单位长度1。

 

解法：利用补图求最短路，由于点比较多，所以利用链式前向星存图，建图后将已存在边标记，然后BFS收缩补图。

 

补图：图G的补图，通俗的来讲就是完全图K_n去除G的边集后得到的图K_n-G。在图论里面，一个图G的补图（complement）或者反面（inverse）是一个图有着跟G相同的点，而且这些点之间有边相连当且仅当在G里面他们没有边相连。

的补图就是

 

对于BFS我们这样编写

void bfs(int start,int n)
{
    set<int> s1;
    set<int> s2;
    set<int>::iterator it;
        for(int i=1;i<=n;i++)
        {
            if(i!=start)
            s1.insert(i);
        }
    dis[start]=0;
    queue<int> que;
    que.push(start);
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].in;
            if(!s1.count(v))continue;
            s1.erase(v);
            s2.insert(v);
        }
        for(it=s1.begin();it!=s1.end();it++)
        {
            que.push(*it);
            dis[*it]=dis[u]+1;
        }
        s1.swap(s2);
        s2.clear();
    }
}

利用两个set容器进行对已求得最短路的点、未求得的点和连接了原图边的点的区分和筛选。

例如对上诉图

先将3和2编入再依次对剩下的点进行搜索。

 

 

 最后得到定点搭配其他点的最短路

全部代码如下

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<stack>
#include<stdlib.h>
#include<map>
#include<vector>
#include<queue>
#include<set>
#include<ctype.h>
#define zuida 100000
using namespace std;
#define INF 0x3f3f3f3f
struct node
{
    int in,next,vil;
}edge[1000000];
int dis[1000000],vis[1000000],head[1000000];
int totl;
void init()
{
    totl=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v,int val)
{
    edge[totl].in=v;
    edge[totl].vil=val;
    edge[totl].next=head[u];
    head[u]=totl++;
}

void bfs(int start,int n)
{
    set<int> s1;
    set<int> s2;
    set<int>::iterator it;
        for(int i=1;i<=n;i++)
        {
            if(i!=start)
            s1.insert(i);
        }
    dis[start]=0;
    queue<int> que;
    que.push(start);
    while(!que.empty())
    {
        int u=que.front();
        que.pop();
        for(int i=head[u];i!=-1;i=edge[i].next)
        {
            int v=edge[i].in;
            if(!s1.count(v))continue;
            s1.erase(v);
            s2.insert(v);
        }
        for(it=s1.begin();it!=s1.end();it++)
        {
            que.push(*it);
            dis[*it]=dis[u]+1;
        }
        s1.swap(s2);
        s2.clear();
    }
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        init();
        int n,m;
        int a,b,c;
        cin>>n>>m;
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&a,&b);
            add(a,b,1);
            add(b,a,1);
        }
        scanf("%d",&c);
        bfs(c,n);
        for(int i=1;i<=n;i++)
        {
            if(i==c)continue;
            if(dis[i]!=INF)printf("%d",dis[i]);
            else printf("-1");
            if(i!=n)printf(" ");
            else printf("\n");
        }
    }
    return 0;
}