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

2
5 5
2 1
4 5
2 3
1 3
3 5
7 2
3 7
4 2

output

3
1 3
3 5
5 4
3 2
2 1
3
2 4

3 7

#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <set> #include <vector> using namespace std; const int N = 210; set<int>q[N]; typedef pair<int,int>node; vector<node>ans; void dfs(int x) {     while(q[x].size())     {         int u=*q[x].begin();         q[x].erase(u), q[u].erase(x);         ans.push_back(make_pair(x,u));         dfs(u);     } } int main() {     int t;     scanf("%d", &t);     while(t--)     {         ans.clear();         int n, m;         scanf("%d %d", &n, &m);         for(int i=0;i<=n;i++)             q[i].clear();         for(int i=0;i<m;i++)         {             int x, y;             scanf("%d %d", &x, &y);             q[x].insert(y);             q[y].insert(x);         }         for(int i=1;i<=n;i++)         {             if(q[i].size()%2==1)             {                 q[n+1].insert(i);                 q[i].insert(n+1);             }         }         printf("%d\n",n-q[n+1].size());         for(int i=1;i<=n;i++)         {             dfs(i);         }         for(int i=0;i<ans.size();i++)         {             if(ans[i].first!=n+1&&ans[i].second!=n+1)             {                 printf("%d %d\n",ans[i].first,ans[i].second);             }         }     }     return 0; }

參考：

題意

給你一個無向圖，然後讓你給邊定向，使得入度等於出度的點最多。

題解：

考慮尤拉圖，只要所有點的度數為偶數就可以了。

那麼答案就是奇數度數的點集，然後我們不停的dfs，把尤拉路都畫出來就行了。

程式碼

#include<bits/stdc++.h>
using namespace std;
const int maxn = 210;
set<int>s[maxn];
vector<pair<int,int> >ans;
int n,m;
void dfs(int x)
{
    while(s[x].size())
    {
        int p=*s[x].begin();
        s[x].erase(p),s[p].erase(x);
        ans.push_back(make_pair(x,p));
        dfs(p);
    }
}
int main()
{
    int t;scanf("%d",&t);
    while(t--)
    {
        ans.clear();
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++)
        {
            int x,y;scanf("%d%d",&x,&y);
            s[x].insert(y);
            s[y].insert(x);
        }
        for(int i=1;i<=n;i++)
        {
            if(s[i].size()%2==1)
                s[n+1].insert(i),s[i].insert(n+1);
        }
        cout<<n-s[n+1].size()<<endl;
        for(int i=1;i<=n;i++)
            dfs(i);
        for(int i=0;i<ans.size();i++)
            if(ans[i].first!=n+1&&ans[i].second!=n+1)
                cout<<ans[i].first<<" "<<ans[i].second<<endl;
    }
}