這題應該算是一道比較簡單的二分圖問題,可是我壓根還是沒想到二分圖,一看解題報告說用二分圖,再一琢磨就明白了。。。。就是求二分圖的最大匹配和關鍵匹配
code:
#include <cstdio>
#include <cstring>
using namespace std;
int n,m,k;
int link[101];
bool graph[101][101],vis[101];
struct pos
{
int x,y;
}p[101*101];
bool find(int x)
{
int y;
for(y=1;y<=m;y++)
{
if(graph[x][y] && !vis[y])
{
vis[y]=true;
if(link[y]==0 || find(link[y]))
{
link[y]=x;
return true;
}
}
}
return false;
}
int solve()
{
int x,ans=0;
memset(link,0,sizeof(link));
for(x=1;x<=n;x++)
{
memset(vis,false,sizeof(vis));
if(find(x))
{
ans++;
}
}
return ans;
}
int main()
{
int i,ans,cnt,cas=0;
while(~scanf("%d%d%d",&n,&m,&k))
{
cnt=0;
memset(graph,false,sizeof(graph));
for(i=1;i<=k;i++)
{
scanf("%d%d",&p[i].x,&p[i].y);
graph[p[i].x][p[i].y]=true;
}
ans=solve();
for(i=1;i<=k;i++)
{
graph[p[i].x][p[i].y]=false;
if(solve()<ans)
{
cnt++;
}
graph[p[i].x][p[i].y]=true;
}
printf("Board %d have %d important blanks for %d chessmen.\n",++cas,cnt,ans);
}
return 0;
}