hdu 5348 MZL's endless loop 暴搜

链接：http://acm.hdu.edu.cn/showproblem.php?pid=5348

MZL's endless loop

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1426    Accepted Submission(s): 319
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

23 31 22 33 17 61 21 31 41 51 61 7

 

Sample Output

111010101

 

题意：给n点m条无向边，要给每条边定一个方向，最后图中所有点的出入度差小于等于1.

做法：暴搜，注意用完的边要删去，更新下head就可以删边了。 然后 每个dfs 只找一条路径， 找到后就break。

#pragma comment(linker, "/STACK:102400000,102400000")  #include <stdio.h>#include <string>#include <iostream>using namespace std;const int N=100100;const int M=300100;struct Edge{int to,  nex , id , fan;}edge[M*2];int head[N],edgenum;//2个要初始化-1和0void add(int u, int v, int id ,int fan){Edge E = { v,  head[u], id, fan};edge[ edgenum ] = E;head[ u ] = edgenum++;}void init(){edgenum=0;}//for(int i=head[nw];i!=-1;i=edge[i].nex)int du[N];//出减  入加int bian[M];void dfs(int nw,int fx){for(int i=head[nw];i!=-1;i=head[nw]){int to=edge[i].to;if(bian[edge[i].id]==-1)//还没确定方向{if(du[nw]!=-1&&du[to]!=1&&fx!=2){ du[nw]--;du[to]++;bian[edge[i].id]=(1^edge[i].fan); head[nw]=edge[i].nex;dfs(to,1);}else if(du[nw]!=1&&du[to]!=-1&&fx!=1){ du[nw]++;du[to]--; bian[edge[i].id]=(0^edge[i].fan); head[nw]=edge[i].nex;dfs(to,2);}elsecontinue; break;} elsehead[nw]=edge[i].nex;//删边 }}int main(){int t;scanf("%d",&t);while(t--){int n,m;init();scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) head[i]=-1;for(int i=0;i<m;i++){bian[i]=-1;int u,v;scanf("%d%d",&u,&v);if(u==v){bian[i]=0;continue;}add(u,v,i,0);add(v,u,i,1);}for(int i=1;i<=n;i++) du[i]=0;for(int i=1;i<=n;i++) {while(head[i]!=-1)dfs(i,0);}for(int i=0;i<m;i++) printf("%d\n",bian[i]);  }return 0;}