[拓扑排序] Jzoj P4269 挑竹签
题解
- 其实这题可以转换为求出有多少个点不在环内
- 容易想到拓扑排序
- 拓扑排序的做法:
- 每次加入没有入度的点,在将与它相连的边删去
- 那么能被删去的点,其实就是不在环内的点
代码
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 #include<cmath> 5 #include<cstring> 6 using namespace std; 7 struct edge{ int to,from; }e[1000010]; 8 int n,m,head,tail,ans,k[1000010],d[1000010],cnt,last[1000010]; 9 void insert(int x,int y) { d[y]++; e[++cnt].to=y; e[cnt].from=last[x]; last[x]=cnt; } 10 int main() 11 { 12 freopen("mikado.in","r",stdin); 13 freopen("mikado.out","w",stdout); 14 scanf("%d%d",&n,&m); 15 for (int i=1;i<=m;i++) 16 { 17 int x,y; 18 scanf("%d%d",&x,&y); 19 insert(x,y); 20 } 21 head=1; tail=0; 22 for (int i=1;i<=n;i++) 23 if (d[i]==0) 24 k[++tail]=i; 25 while (head<=tail) 26 { 27 int u=k[head]; head++; 28 ans++; 29 for (int i=last[u];i;i=e[i].from) 30 { 31 d[e[i].to]--; 32 if (d[e[i].to]==0) k[++tail]=e[i].to; 33 } 34 } 35 printf("%d",ans); 36 return 0; 37 }