CF 723E 巧用欧拉回路

E. One-Way Reform
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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：给200个点的无向图，不一定连通，现在求一种标方向的方案，使得入度等于出度的点最多。

做法：显然最好情况是每个度数为偶的节点都使入度等于出度，感觉确实可以做到，下面构造一种方案。首先，对于每一个连通块，度数为奇数的点有偶数个，不妨将入度为奇数的点两两连一条虚线，这样每个连通块都是一个欧拉回路。dfs跑一遍欧拉回路路径，虚线不输出即可。因为虚线只影响奇点的奇偶性，而奇点肯定不对，所以对最终结果没有影响。写的时候注意由于虚线可能和实线重复，标记+2即可。

代码：

#include <bits/stdc++.h>using namespace std;int g[205][205];int degree[205];int vis[205];int n, m;int tot;void dfs(int st, int tag){    vis[st]=tag;    for(int i=1;i<=n;i++)    {        if(g[st][i]&&!vis[i])dfs(i, tag);    }}void dfs2(int st){    for(int i=1;i<=n;i++)    {        if(g[st][i]==1){            g[st][i]=g[i][st]=0;            dfs2(i);            printf("%d %d\n", i, st);        }        else if(g[st][i]>=2){            g[st][i]-=2;            g[i][st]-=2;            dfs2(i);        }    }}int main(){    int t;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &n, &m);        memset(degree, 0, sizeof(degree));        memset(g, 0, sizeof(g));        memset(vis, 0, sizeof(vis));        for(int i=1;i<=m;i++){            int x, y;            scanf("%d%d", &x, &y);            degree[x]++;            degree[y]++;            g[x][y]++;            g[y][x]++;        }        int cnt=0;        for(int i=1;i<=n;i++)if(degree[i]%2==0)cnt++;        printf("%d\n", cnt);        int last=-1;        tot=0;        for(int i=1;i<=n;i++){            if(!vis[i])dfs(i, ++tot);        }        //printf("%d\n", tot);        for(int i=1;i<=tot;i++){            for(int j=1;j<=n;j++){                if(vis[j]==i&&degree[j]%2){                    if(last!=-1){                        g[j][last]+=2;                        g[last][j]+=2;                        last=-1;                    }                    else{                        last=j;                    }                }            }            for(int j=1;j<=n;j++){                if(vis[j]==i){                    dfs2(j);                }            }        }    }}