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 ;
}