codeforces 723E. One-Way Reform(欧拉回路||网络流)

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

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

题意：

有一个无向图，n个点，m条边，没有自环没有重边，将无向图变成有向图，给每条边定方向，使得最终入度等于出度的点的个数最多。

输出入度等于出度的点的个数，以及这m条有向边。

思路：

【先求无向图中每个点的度数，每次都从度数为奇数的点开始走，删掉走过的边，记录每个点的入度跟出度，一直走到不能走为止。恩 这样做有一个明显的错误 当输入的无向图是一个环的时候 就没有所谓的度数为奇数的点 因此在把度数为奇数的点处理完了之后还要检查是否存在偶数的点 (⊙o⊙)…】

#include <bits/stdc++.h>  
using namespace std;  
#define ll __int64  
const int N=1e6+10;  
  
int n, m;  
int line[250][250];  
vector<int>vec[250], vv[250];  
int ru[250], chu[250], du[250];  
queue<int>q;  

void solve(){
	while(!q.empty()){  
        int x = q.front();q.pop();  
        while(1){  
            int len = vec[x].size();  
            int flag = 0;  
            for(int i=0; i<len; i++){  
                int v = vec[x][i];  
                if(line[x][v]){  
                    line[x][v]--, line[v][x]--;  
                    du[x]--, du[v]--;  
                    chu[x]++, ru[v]++;  
                    vv[x].push_back(v);  
                    if(du[x]&1) q.push(x);  
                    if(du[v]&1) q.push(v);  
                    x = v;  
                    flag=1;  
                    break;  
                }  
            }  
            if(flag==0) break;  
        }  
    }  
}

int main(){  
    int t;  
    scanf("%d", &t);  
    while(t--){  
        memset(line, 0, sizeof(line));  
        scanf("%d%d", &n, &m);  
        for(int i=1; i<=n; i++) vec[i].clear(), vv[i].clear();  
        int u, v;  
        memset(du, 0, sizeof(du));  
        for(int i=0; i<m; i++){  
            scanf("%d%d", &u, &v);  
            line[u][v]++;  
            line[v][u]++;  
            vec[u].push_back(v);  
            vec[v].push_back(u);  
            du[u]++, du[v]++;  
        }  
  
        memset(ru, 0, sizeof(ru));  
        memset(chu, 0, sizeof(chu));  
        while(!q.empty()) q.pop();  
          
        for(int i=1; i<=n; i++){  
            if(du[i]&1) q.push(i);  
        }  
        solve();
        for(int i=1; i<=n; i++){  
            if(du[i]) q.push(i);  
        }  
        solve();
        int ans=0;  
        for(int i=1; i<=n; i++){  
            if(ru[i] == chu[i]) ans++;  
        }  
        printf("%d\n", ans);  
        for(int i=1; i<=n; i++){  
            int len = (int)vv[i].size();  
            for(int j=0; j<len; j++){  
                printf("%d %d\n", i, vv[i][j]);  
            }  
        }  
    }  
    return 0;  
}