http://acm.hdu.edu.cn/showproblem.php?pid=3018
Ant Trip
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3794 Accepted Submission(s): 1502
Problem Description
Ant Country consist of N towns.There are M roads connecting the towns.
Ant Tony,together with his friends,wants to go through every part of the country.
They intend to visit every road , and every road must be visited for exact one time.However,it may be a mission impossible for only one group of people.So they are trying to divide all the people into several groups,and each may start at different town.Now tony wants to know what is the least groups of ants that needs to form to achieve their goal.
Ant Tony,together with his friends,wants to go through every part of the country.
They intend to visit every road , and every road must be visited for exact one time.However,it may be a mission impossible for only one group of people.So they are trying to divide all the people into several groups,and each may start at different town.Now tony wants to know what is the least groups of ants that needs to form to achieve their goal.
Input
Input contains multiple cases.Test cases are separated by several blank lines. Each test case starts with two integer N(1<=N<=100000),M(0<=M<=200000),indicating that there are N towns and M roads in Ant Country.Followed by M lines,each line contains two integers a,b,(1<=a,b<=N) indicating that there is a road connecting town a and town b.No two roads will be the same,and there is no road connecting the same town.
Output
For each test case ,output the least groups that needs to form to achieve their goal.
Sample Input
3 3
1 2
2 3
1 3
4 2
1 2
3 4
Sample Output
1
2
题目大意：给一个图【不保证连通】，求最少需要几画才能在不重复使用边的情况下走完全部边。
题目分析：一笔画模板题。
需要画数=每个连通图的画数之和。 每个连通图的画数=如果各个点的度全部是大于0的偶数则根据欧拉回路为1，如果存在奇数，则为奇数个数/2。
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<queue> 5 #include<vector> 6 using namespace std; 7 int in[100005]; 8 int pre[100005]; 9 int tot1[100005],tot2[100005]; 10 int n,m; 11 int find(int x) 12 { 13 int xx=x; 14 while(x!=pre[x]) 15 { 16 x=pre[x]; 17 } 18 while(pre[xx]!=x) 19 { 20 int t=pre[xx]; 21 pre[xx]=x; 22 xx=t; 23 } 24 return x; 25 } 26 int main() 27 { 28 while(scanf("%d%d",&n,&m)==2) 29 { 30 memset(tot1,0,sizeof(tot1)); 31 memset(tot2,0,sizeof(tot2)); 32 memset(in,0,sizeof(in)); 33 for(int i = 1 ; i <= n ; i++) 34 { 35 pre[i]=i; 36 } 37 memset(in,0,sizeof(in)); 38 while(m--) 39 { 40 int a,b; 41 scanf("%d%d",&a,&b); 42 in[a]++; 43 in[b]++; 44 if(find(a)!=find(b)) 45 { 46 pre[find(a)]=find(b); 47 } 48 } 49 for(int i = 1 ; i <= n ; i++) 50 { 51 int q=find(i); 52 if(in[i]%2==0&&in[i]) 53 tot1[q]++; 54 else if(in[i]%2) 55 tot2[q]++; 56 } 57 int sum=0; 58 for(int i = 1 ; i <= n ;i++) 59 { 60 if(tot2[i]) 61 { 62 sum+=tot2[i]/2; 63 } 64 else if(tot1[i]) 65 { 66 sum++; 67 } 68 } 69 cout << sum << endl; 70 71 } 72 return 0; 73 }