可以发现,题目求得就是一个可相交最小路径覆盖,但是有环。所以我们先用tarjan把环都缩掉,然后Floyd传递闭包,求二分图最大匹配,答案就是scc-ans.
#include#include #include #include using namespace std; #define ll long long #define N 210 #define M 1010 inline int read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar(); return x*f; } int tst,n,m,h[N],h1[N],num=1,scc=0,bel[N],dfn[N],low[N],dfnum=0,ans=0,bf[N]; bool inq[N],mp[N][N]; struct edge{ int to,next; }data1[M],data[M<<2]; stack qq; inline void add(int x,int y){ data[++num].to=y;data[num].next=h[x];h[x]=num; } inline void tarjan(int x){ dfn[x]=low[x]=++dfnum;qq.push(x);inq[x]=1; for(int i=h1[x];i;i=data1[i].next){ int y=data1[i].to; if(!dfn[y]) tarjan(y),low[x]=min(low[x],low[y]); else if(inq[y]) low[x]=min(low[x],dfn[y]); }if(dfn[x]==low[x]){ ++scc;while(1){ int y=qq.top();qq.pop();inq[y]=0; bel[y]=scc;if(y==x) break; } } } inline bool find(int x){ for(int i=h[x];i;i=data[i].next){ int y=data[i].to;if(inq[y]) continue; inq[y]=1;if(!bf[y]||find(bf[y])){bf[y]=x;return 1;} }return 0; } int main(){ // freopen("a.in","r",stdin); tst=read(); while(tst--){ n=read();m=read();memset(h1,0,sizeof(h1));memset(h,0,sizeof(h)); num=1;scc=0;memset(dfn,0,sizeof(dfn));dfnum=0;memset(mp,0,sizeof(mp)); memset(inq,0,sizeof(inq));ans=0;memset(bf,0,sizeof(bf)); while(m--){ int x=read(),y=read(); data1[++num].to=y;data1[num].next=h1[x];h1[x]=num; }for(int i=1;i<=n;++i) if(!dfn[i]) tarjan(i);num=1; for(int x=1;x<=n;++x) for(int i=h1[x];i;i=data1[i].next){ int y=data1[i].to;if(bel[y]!=bel[x]) mp[bel[x]][bel[y]]=1; } for(int k=1;k<=scc;++k) for(int i=1;i<=scc;++i) for(int j=1;j<=scc;++j) mp[i][j]|=(mp[i][k]&mp[k][j]); for(int i=1;i<=scc;++i) for(int j=1;j<=scc;++j) if(mp[i][j]) add(i,j); for(int i=1;i<=scc;++i){ memset(inq,0,sizeof(inq));if(find(i)) ans++; }printf("%d\n",scc-ans); } return 0; }