C. Graph Reconstruction
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/329/problem/C
Description
I have an undirected graph consisting of n nodes, numbered 1 through n. Each node has at most two incident edges. For each pair of nodes, there is at most an edge connecting them. No edge connects a node to itself.
I would like to create a new graph in such a way that:
- The new graph consists of the same number of nodes and edges as the old graph.
- The properties in the first paragraph still hold.
- For each two nodes u and v, if there is an edge connecting them in the old graph, there is no edge connecting them in the new graph.
Help me construct the new graph, or tell me if it is impossible.
⋅1. If you touch a buoy before your opponent, you will get one point. For example if your opponent touch the buoy #2 before you after start, he will score one point. So when you touch the buoy #2, you won't get any point. Meanwhile, you cannot touch buoy #3 or any other buoys before touching the buoy #2.
⋅2. Ignoring the buoys and relying on dogfighting to get point.
If you and your opponent meet in the same position, you can try to
fight with your opponent to score one point. For the proposal of game
balance, two players are not allowed to fight before buoy #2 is touched by anybody.
There are three types of players.
Speeder:
As a player specializing in high speed movement, he/she tries to avoid
dogfighting while attempting to gain points by touching buoys.
Fighter:
As a player specializing in dogfighting, he/she always tries to fight
with the opponent to score points. Since a fighter is slower than a
speeder, it's difficult for him/her to score points by touching buoys
when the opponent is a speeder.
All-Rounder: A balanced player between Fighter and Speeder.
There will be a training match between Asuka (All-Rounder) and Shion (Speeder).
Since the match is only a training match, the rules are simplified: the game will end after the buoy #1 is touched by anybody. Shion is a speed lover, and his strategy is very simple: touch buoy #2,#3,#4,#1 along the shortest path.
Asuka is good at dogfighting, so she will always score one point by dogfighting with Shion, and the opponent will be stunned for T seconds after dogfighting.
Since Asuka is slower than Shion, she decides to fight with Shion for
only one time during the match. It is also assumed that if Asuka and
Shion touch the buoy in the same time, the point will be given to Asuka
and Asuka could also fight with Shion at the buoy. We assume that in
such scenario, the dogfighting must happen after the buoy is touched by
Asuka or Shion.
The speed of Asuka is V1 m/s. The speed of Shion is V2 m/s. Is there any possibility for Asuka to win the match (to have higher score)?
Input
The first line consists of two space-separated integers: n and m (1 ≤ m ≤ n ≤ 105), denoting the number of nodes and edges, respectively. Then m lines follow. Each of the m lines consists of two space-separated integers u and v (1 ≤ u, v ≤ n; u ≠ v), denoting an edge between nodes u and v.
Output
If it is not possible to construct a new graph with the mentioned properties, output a single line consisting of -1. Otherwise, output exactly m lines. Each line should contain a description of edge in the same way as used in the input format.
Sample Input
8 7
1 2
2 3
4 5
5 6
6 8
8 7
7 4
Sample Output
1 4
4 6
1 6
2 7
7 5
8 5
2 8
HINT
题意
给你一个n个点m条边的无向图
无自环，无重边，每个点的度数最多为2
然后让你找到一个图，使得性质一样，但是之前相连的边，之后不能相连
题解：
随机化，首先满足题意的图应该有很多个，所以瞎随……
注意随机的时候，用一种保证每个点的度数最多为2的方式随机就好了
代码
#include<iostream>
#include<stdio.h>
#include<map>
#include<vector>
#include<algorithm>
#include<ctime>
using namespace std; map<pair<int,int> ,int>H;
int main()
{
srand(time(NULL));
int n,m;
scanf("%d%d",&n,&m);
vector<int> Q;
for(int i=;i<=n;i++)
Q.push_back(i);
for(int i=;i<=m;i++)
{
int x,y;scanf("%d%d",&x,&y);
if(x>y)swap(x,y);
H[make_pair(x,y)]=;
}
int tot = ;
while()
{
tot++;
random_shuffle(Q.begin(),Q.end());
int flag = ;
for(int i=;i<m;i++)
{
int k = Q[i],p = Q[(i+)%n];
if(k==p)
{
flag=;
break;
}
if(k>p)
swap(k,p);
if(H[make_pair(k,p)])
{
flag=;
break;
}
}
if(flag==)
break;
if(tot==)
{
printf("-1\n");
return ;
}
}
for(int i=;i<m;i++)
printf("%d %d\n",Q[(i+)%n],Q[i]); }