inpu print string app oom set vid ans 2個

There are a group of students. Some of them may know each other, while others don‘t. For example, A and B know each other, B and C know each other. But this may not imply that A and C know each other.

Now you are given all pairs of students who know each other. Your task is to divide the students into two groups so that any two students in the same group don‘t know each other.If this goal can be achieved, then arrange them into double rooms. Remember, only paris appearing in the previous given set can live in the same room, which means only known students can live in the same room.

Calculate the maximum number of pairs that can be arranged into these double rooms.

InputFor each data set:
The first line gives two integers, n and m(1<n<=200), indicating there are n students and m pairs of students who know each other. The next m lines give such pairs.

Proceed to the end of file.

OutputIf these students cannot be divided into two groups, print "No". Otherwise, print the maximum number of pairs that can be arranged in those rooms.
Sample Input

4 4
1 2
1 3
1 4
2 3
6 5
1 2
1 3
1 4
2 5
3 6

Sample Output

No
3

點染色判定二分圖，將二分圖分成2個部分，之後匈牙利求最大匹配即可。

 1 #include<iostream>
 2
 
 using namespace std;
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<vector>
 6 const int maxn = 1000;
 7 vector <int> maps[maxn];
 8 int n,m;
 9 int ok[maxn];//ok[i]代表第i個節點可否配對 
10 int matched[maxn];//matched[i]代表第i個節點的配對對象; 
11 bool match(int x){
12     for(int i=0;i<maps[x].size();i++){
13         int k = maps[x][i];
14         if(ok[k]==0){
15             ok[k]=1;
16             if(matched[k]==0||match(matched[k])==true){
17                 matched[k]=x;
18                 return true;
19             }
20         }
21     }
22     return false;
23 }
24 int ans = 1;
25 int color[1100];
26 void dfs(int x,int Color){
27     color[x]=Color;
28     for(int i=0;i<maps[x].size()&&ans;i++){
29         if(color[maps[x][i]]!=-1){
30             if(color[maps[x][i]]==Color){
31                 ans = 0;
32                 return;
33             }
34         }
35         else
36         dfs(maps[x][i],!Color);
37     }
38 }
39 int main(){
40     while(scanf("%d%d",&n,&m)!=EOF){
41         for(int i=1;i<=n;i++)
42             maps[i].clear();
43         memset(matched,0,sizeof(matched));
44         ans = 1;
45         for(int i=0;i<n;i++)
46             maps[i].clear();
47         memset(color,-1,sizeof(color));
48         for(int i=1;i<=m;i++){
49             int k,q;
50             scanf("%d%d",&k,&q);
51             maps[k].push_back(q);
52             maps[q].push_back(k);
53         }
54         ans = 1;
55         for(int i=1;i<=n&&ans;i++){
56             if(color[i]!=-1)
57                 continue;
58             dfs(i,0);
59         }
60         if(!ans){
61             printf("No\n");
62             continue;
63         }
64         ans = 0;
65         for(int i=1;i<=n;i++){
66             memset(ok,0,sizeof(ok));
67             if(color[i]!=0)
68                 continue;
69             if(match(i)==true)
70                 ans+=1;
71         }
72         cout<<ans<<endl;
73     }
74     return 0;
75 }

HDU - 2444 The Accomodation of Studentsp[二分圖判定,匈牙利算法]