hdu5348 MZL's endless loop(欧拉回路)

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MZL's endless loop

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1705    Accepted Submission(s): 369
Special Judge

Problem Description

As we all kown, MZL hates the endless loop deeply, and he commands you to solve this problem to end the loop.
You are given an undirected graph with n vertexs and m edges. Please direct all the edges so that for every vertex in the graph the inequation |out degree − in degree|≤1 is satisified.
The graph you are given maybe contains self loops or multiple edges.

 

Input

The first line of the input is a single integer T, indicating the number of testcases.
For each test case, the first line contains two integers n and m.
And the next m lines, each line contains two integers ui and vi, which describe an edge of the graph.
T≤100, 1≤n≤105, 1≤m≤3∗105, ∑n≤2∗105, ∑m≤7∗105.

 

Output

For each test case, if there is no solution, print a single line with −1, otherwise output m lines,.
In ith line contains a integer 1 or 0, 1 for direct the ith edge to ui→vi, 0 for ui←vi.

 

Sample Input

2

3 3

1 2

2 3

3 1

7 6

1 2

1 3

1 4

1 5

1 6

1 7

 

Sample Output

1

1

1

0

1

0

1

0

1

 

题意就是给你一张无向图，让你把它变成有向图，使得对于每一个顶点都满足出度与入度的差的绝对值小于等于一

利用欧拉回路，在欧拉图中，每个点的出度都等于入度，那么对于这个图，其实就相当于若干个欧拉图，然后去掉一些边。

然后我们需要做的就是补边。也就是对于每个奇度点，加一条连到其它奇度点的无向边，然后跑欧拉回路，跑的方向就是这条边的方向。

另外注意有多个连通分支。这题比较容易T，虽然我的队友在比赛时瞬间就AC了。。。然而我还是在赛后T了好久，毕竟队友是final选手

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author xyiyy @https://github.com/xyiyy
 */

#include <iostream>
#include <fstream>

//#####################
//Author:fraud
//Blog: http://www.cnblogs.com/fraud/
//#####################
#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <sstream>
#include <ios>
#include <iomanip>
#include <functional>
#include <algorithm>
#include <vector>
#include <string>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <climits>
#include <cctype>

using namespace std;
#define rep(X, N) for(int X=0;X<N;X++)

const int MAXN = 800010;
int head[MAXN];
int Next[MAXN], To[MAXN];
int vis[MAXN];
int used[100010];
int deg[100010];
int gao;
int tot;

void init(int n) {
    tot = 0;
    rep(i, n)head[i] = -1;
}

void addedge(int u, int v) {
    Next[tot] = head[u];
    To[tot] = v;
    vis[tot] = 0;
    head[u] = tot++;
}

void eular(int u){
    used[u] = 1;
    int i;
    while(head[u]!=-1){
    //for(int &i = head[u];i != -1;i = Next[i]){
        i = head[u];
        head[u] = Next[head[u]];
        if(vis[i])continue;
        vis[i^1] = 1;
        eular(To[i]);
    }
}
int Scan() {
    int res=0, ch;
    while(ch=getchar(), ch<'0'||ch>'9');
    res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')
        res=res*10+ch-'0';
    return res;
}
void Out(int a) {
    if (a > 9)
        Out(a / 10);
    putchar(a % 10 + '0');
}

class hdu5348 {
public:
    void solve() {
        int t;
        t =Scan();//in >> t;
        while (t--) {
            int n, m;
            n = Scan();m=Scan();//in >> n >> m;
            init(n);
            rep(i, n)deg[i] = 0;
            int u, v;
            rep(i, m) {
                u = Scan();v= Scan();//in >> u >> v;
                u--, v--;
                deg[u]++;
                deg[v]++;
                addedge(u, v);
                addedge(v, u);
            }
            gao = -1;
            rep(i, n) {
                if (deg[i] & 1) {
                    if (gao != -1) {
                        addedge(i, gao);
                        addedge(gao, i);
                        gao = -1;
                    } else gao = i;
                }
            }
            rep(i, n) used[i] = 0;
            /*rep(i,n){
                if(!used[i]){
                    dfs(i);
                    num++;
                }
            }*/
            gao = -1;
            rep(i, n) {
                if (!used[i]) {
                    eular(i);
                }
            }
            m<<=1;
            for(int i=1;i<m;i+=2){
                if (vis[i])putchar('1');
                else putchar('0');
                putchar('\n');
            }

        }
    }
};


int main() {
    //std::ios::sync_with_stdio(false);
    //std::cin.tie(0);
    hdu5348 solver;
    //std::istream &in(std::cin);
    //std::ostream &out(std::cout);
    solver.solve();
    return 0;
}