题解:
对于每一个孩子裂点+建边
如果i孩子讨厌的和j孩子喜欢的相同,那么建边
然后跑最大独立集
代码:
#include<cstdio> #include<cstring> #include<cmath> #include<algorithm> using namespace std; const int N=2005; int fi[N],num,s1[N],s2[N],match[N],zz[N*N],ne[N*N],n,f[N]; void jb(int x,int y) { ne[++num]=fi[x]; fi[x]=num; zz[num]=y; } int dfs(int x) { for (int i=fi[x];i;i=ne[i]) if (!f[zz[i]]) { f[zz[i]]=1; if (!match[zz[i]]||dfs(match[zz[i]])) { match[zz[i]]=x; return 1; } } return 0; } int read() { int k=0,x=0;char c; for (c=getchar();c!='C'&&c!='D';c=getchar()); if (c=='C')k+=n; scanf("%d",&x); return k+x; } int main() { while (~scanf("%d%d%d",&n,&n,&n)) { memset(fi,0,sizeof fi); memset(match,0,sizeof match); num=0; for (int i=1;i<=n;i++)s1[i]=read(),s2[i]=read(); for (int i=1;i<=n;i++) for (int j=1;j<=n;j++) if (s1[i]==s2[j]||s2[i]==s1[j])jb(i,j); int ans=0; for (int i=1;i<=n;i++) { memset(f,0,sizeof f); ans+=dfs(i); } printf("%d ",n-ans/2); } }