E. One-Way Reform
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
There are n cities and m two-way roads in Berland, each road connects two cities. It is known that there is no more than one road connecting each pair of cities, and there is no road which connects the city with itself. It is possible that there is no way to get from one city to some other city using only these roads.
The road minister decided to make a reform in Berland and to orient all roads in the country, i.e. to make each road one-way. The minister wants to maximize the number of cities, for which the number of roads that begins in the city equals to the number of roads that ends in it.
Input
The first line contains a positive integer t (1 ≤ t ≤ 200) — the number of testsets in the input.
Each of the testsets is given in the following way. The first line contains two integers n and m (1 ≤ n ≤ 200, 0 ≤ m ≤ n·(n - 1) / 2) — the number of cities and the number of roads in Berland.
The next m lines contain the description of roads in Berland. Each line contains two integers u and v (1 ≤ u, v ≤ n) — the cities the corresponding road connects. It's guaranteed that there are no self-loops and multiple roads. It is possible that there is no way along roads between a pair of cities.
It is guaranteed that the total number of cities in all testset of input data doesn't exceed 200.
Pay attention that for hacks, you can only use tests consisting of one testset, so t should be equal to one.
Output
For each testset print the maximum number of such cities that the number of roads that begins in the city, is equal to the number of roads that ends in it.
In the next m lines print oriented roads. First print the number of the city where the road begins and then the number of the city where the road ends. If there are several answers, print any of them. It is allowed to print roads in each test in arbitrary order. Each road should be printed exactly once.
Example
input
Copy
2 5 5 2 1 4 5 2 3 1 3 3 5 7 2 3 7 4 2
output
Copy
3 1 3 3 5 5 4 3 2 2 1 3 2 4 3 7
题意：给一个无向图，现在需要把所有的边变为有向，是的出度与入度相同的点的个数最多。
可以看出度数为偶数的点一定可以满足，重要的是如何输出路径，一个巧妙的方法是将所有度数为奇数的点和第n+1个点建无向边，这样是的所有的点度数都是偶数（度数为奇数的点的个数一定为偶数），然后跑一边欧拉回路，只输出原始边就行了。
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<set>
#include<vector>
using namespace std;
const int maxm = 2005;
int a[maxm], flag[maxm], n;
set<int>s[maxm];
set<int>::iterator it;
void dfs(int k)
{
while (!s[k].empty())
{
int v = *s[k].begin();
s[k].erase(v), s[v].erase(k);
if(k!=n+1&&v!=n+1)
printf("%d %d\n", k, v);
dfs(v);
}
}
int main()
{
int i, j, k, sum, t, m, u, v, ans = 0;
scanf("%d", &t);
while (t--)
{
scanf("%d%d", &n, &m);
for (i = 1;i <= n + 1;i++)
s[i].clear(), flag[i] = 0;
for (i = 1;i <= m;i++)
{
scanf("%d%d", &u, &v);
s[u].insert(v);
s[v].insert(u);
flag[u]++, flag[v]++;
}
ans = 0;
for (i = 1;i <= n;i++)
{
if (flag[i] % 2 != 0)
{
s[i].insert(n + 1);
s[n + 1].insert(i);
ans++;
}
}
printf("%d\n", n - ans);
for (i = 1;i <= n;i++)
dfs(i);
}
}