One-stroke drawing problem
Time limit: 3000 ms | Memory limit: 65535 KB
Difficulty: 4
Description
zyc likes to play some small games since he was a child, including drawing a one-stroke drawing. He wants to ask you to help him write a program to determine whether a picture can be Draw with a single stroke.
It is stipulated that all edges can only be drawn once and cannot be drawn repeatedly.
The first line of input has only one positive integer N (N<=10) representing the number of groups of test data.
The first line of each set of test data has two positive integers P, Q (P<=1000, Q<=2000), which respectively indicate how many vertices and how many lines there are in the painting. (points are numbered from 1 to P)
subsequent Q lines, each with two positive integers A, B (0
/*
判断一笔画的问题
首先图必须保证是联通的
第二是 如果存在一个欧拉路的话 度数为奇数个的点必须为0 或者 2
如果存在一个欧拉回路的话 所有的度数必须是偶数
http://acm.nyist.edu.cn/JudgeOnline/problem.php?pid=42
*/
#include<iostream>
#include<stdio.h>
#include<string.h>
using namespace std;
int du[1007];
int pre[1007];
int vis[1007][1007];//判断边是否重复
int getF(int x)
{
if(x != pre[x]) return pre[x] = getF(pre[x]);
return x;
}
void Union(int x,int y)
{
int pa = getF(x);
int pb = getF(y);
if(pa != pb)
{
pre[pa] = pb;
}
}
int main()
{
int ncase;
scanf("%d",&ncase);
while(ncase--)
{
memset(vis,0,sizeof(vis));
memset(du,0,sizeof(vis));
int n,m;
scanf("%d%d",&n,&m);
for(int i =1;i<= n;i++)
pre[i] = i;
for(int i = 0;i<m;i++)
{
int a,b;
scanf("%d%d",&a,&b);
if(!vis[a][b])//消掉重复边
{
vis[a][b] = 1;
du[a]++;
du[b]++;
Union(a,b);
}
}
int cnt =0;
int flag = 0;
for(int i = 1;i<=n;i++)
{
int f = getF(i);
if(f == i) flag ++;
if(du[i] % 2 ) cnt++;
}
if(flag >= 2)
{
printf("No\n");
}
else
{
if(flag==0)//存在一个回路
{
if(cnt == 0)
printf("Yes\n");
else
printf("No\n");
}
else // 不存在回路
{
if(cnt == 0 || cnt == 2)
{
printf("Yes\n");
}
else
printf("No\n");
}
}
}
return 0;
}