HDU 5876：Sparse Graph（BFS）

http://acm.hdu.edu.cn/showproblem.php?pid=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

题意：给出一个图，和一个起点，求在该图的补图中从起点到其他N-1个点的最短距离。如果不连通输出-1.

思路：比赛的时候觉得这题N那么大不会做。现在回过头发现貌似不难。用set将未走过的点放置进去，并在对点的邻边进行扩展的时候，把能走到的邻点删除掉（即补图中可以走到的邻点保留）。

 #include <cstdio>
 #include <algorithm>
 #include <iostream>
 #include <cstring>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <vector>
 #include <set>
 using namespace std;
 #define N 200010
 #define INF 0x3f3f3f3f

 struct node
 {
     int v, nxt;
 }edge[N];
 int head[N];
 int d[N], tot;

 void add(int u, int v)
 {
     edge[tot].v = v; edge[tot].nxt = head[u]; head[u] = tot++;
     edge[tot].v = u; edge[tot].nxt = head[v]; head[v] = tot++;
 }

 void bfs(int st, int n)
 {
     queue<int> que;
     d[st] = ;
     que.push(st);
     set<int> s1, s2;
     for(int i = ; i <= n; i++) {
         if(i != st) s1.insert(i);
     }
     while(!que.empty()) {
         int u = que.front(); que.pop();
         for(int k = head[u]; ~k; k = edge[k].nxt) {
             int v = edge[k].v;
             if(!s1.count(v)) continue;
             s1.erase(v); //补图中当前能走到的并且还未更新过的
             s2.insert(v); //补图中走不到的还要扩展的
         }
         for(set<int>:: iterator it = s1.begin(); it != s1.end(); it++) {
             d[*it] = d[u] + ;
             que.push(*it);
         }
         s1.swap(s2); //还没更新过的
         s2.clear();
     }
 }

 int main()
 {
     int t;
     scanf("%d", &t);
     while(t--) {
         memset(head, -, sizeof(head));
         memset(d, INF, sizeof(INF));
         int n, m;
         scanf("%d%d", &n, &m);
         for(int i = ; i < m; i++) {
             int u, v;
             scanf("%d%d", &u, &v);
             add(u, v);
         }
         int st;
         scanf("%d", &st);
         bfs(st, n);
         bool f = ;
         for(int i = ; i <= n; i++) {
             if(i != st) {
                 if(f) printf(" ");
                 f = ;
                 if(d[i] == INF) printf("-1");
                 else printf("%d", d[i]);
             }
         }
         puts("");
     }

     return ;
 }