Description
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.
Input
For 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.
Output
If 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
Source
2008 Asia Harbin Regional Contest Online
最近一直沒寫blog
發現二分圖還是不太會做
這個題是先判斷是不是二分圖，然后求最大匹配
題意
有n個學生，有m對人是認識的，每一對認識的人能分到一間房，問能否把n個學生分成兩
部分，每部分內的學生互不認識，而兩部分之間的學生認識。如果可以分成兩部分，就
算出房間最多需要多少間，否則就輸出No。
#include<stdio.h>
#include<string.h>
#include<math.h>
#define maxn 205
int n,m;
int g[maxn][maxn];
int color[maxn];
int cx[maxn],cy[maxn];
int mk[maxn];
int ans;
bool flag;
int path(int u)
{
int v;
for(v=1; v<=n; v++)
{
if(g[u][v]&&!mk[v])
{
mk[v]=1;
if(cy[v]==-1||path(cy[v]))
{
cx[u]=v;
cy[v]=u;
return 1;
}
}
}
return 0;
}
int match()
{
int res,i;
res=0;
memset(cx,-1,sizeof(cx));
memset(cy,-1,sizeof(cy));
for(i=1; i<=n; i++)
if(cx[i]==-1)
{
memset(mk,0,sizeof(mk));
res+=path(i);
}
return res;
}
void dfs(int u,int col)
{
int i;
color[u]=col;
for(i=1;i<n;i++)
if((g[u][i]||g[i][u])&&color[i]==-1&&flag)
{
dfs(i,1-col);
}
else if((g[u][i]||g[i][u])&&color[i]==col)
{
flag=false;
return;
}
}
int main()
{
int i,j;
int p1,p2;
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(g,0,sizeof(g));
for(i=1; i<=m; i++)
{
scanf("%d%d",&p1,&p2);
g[p1][p2]=1;
}
ans=0;
flag=true;
memset(color,-1,sizeof(color));
dfs(1,0);
if (!flag) printf("No\n");
else
{
ans=match();
printf("%d\n",ans);
}
}
return 0;
}
//二分圖判斷+最大匹配