题意:有一群牛,求被所有牛都认可的牛的个数
每个连通分量建一个缩点,出度为零的缩点包含的点的个数即为要求值
如果有多个出度为零的,直接输出零,否则输出那唯一一个出度为零的缩点包含的点的个数
#include<stdio.h> #include<string.h> #define N 11000 int dfn[N],low[N],sta[N],visit[N],suo[N],ans,outdegree[N],top,num[N]; int head[N],yong,n,m; struct node { int v,next; }bian[N*10]; void init() { memset(sta,0,sizeof(sta)); ans=0; memset(outdegree,0,sizeof(outdegree)); memset(suo,0,sizeof(suo)); memset(dfn,0,sizeof(dfn)); memset(low,0,sizeof(low)); memset(visit,0,sizeof(visit));yong=0;top=0; memset(head,-1,sizeof(head)); memset(num,0,sizeof(num)); } void addedge(int u,int v){ bian[yong].v=v; bian[yong].next=head[u]; head[u]=yong++; } int Min(int u,int v) { return u>v?v:u; } void tarjan(int u) { visit[u]=1; dfn[u]=++yong; low[u]=yong; sta[++top]=u; int i,v; for(i=head[u];i!=-1;i=bian[i].next) { v=bian[i].v; if(!dfn[v]) { tarjan(v); low[u]=Min(low[u],low[v]); } else if(visit[v]==1) low[u]=Min(low[u],dfn[v]); } int coun=0; if(dfn[u]==low[u]) { ans++; do{ v=sta[top--]; coun++; visit[v]=2; suo[v]=ans;//缩点 }while(v!=u); num[ans]=coun;//记录缩点包含的点的个数 } } int main() { int i,j,a,b; while(scanf("%d%d",&n,&m)!=EOF) { init(); while(m--) { scanf("%d%d",&a,&b); addedge(a,b); } ans=0; yong=0; for(i=1;i<=n;i++) if(visit[i]!=2) tarjan(i); /* for(i=1;i<=n;i++) printf("%d ",suo[i]);*/ for(i=1;i<=n;i++) for(j=head[i];j!=-1;j=bian[j].next) if(suo[i]!=suo[bian[j].v]) outdegree[suo[i]]++; yong=0; for(i=1;i<=ans;i++) if(outdegree[i]==0){ yong++; j=i; } if(yong>1) printf("0 "); else printf("%d ",num[j]); } return 0; }