「HNOI2015」菜肴制作

「HNOI2015」菜肴制作

这道题想到了其实还挺水的，一开始我直接用小根堆拓扑然后就爆0了，然后我又用十万个堆搜索，T30，还是xkl告诉我要倒着拓扑。

首先要建反图，对于入度为0的点，较小的点先输出所以要优先拓扑大的点，这样就保证了大的点及其子树（其实并不是树，这样好理解点）都存在数组前面，再倒着输出即可。

#include<iostream>
#include<cstring>
#include<cstdio>
#include<time.h>
#include<queue>
#define ma(x) memset(x,0,sizeof(x))
using namespace std;
struct edge
{
	int u,v,nxt;
	#define u(x) ed[x].u
	#define v(x) ed[x].v
	#define n(x) ed[x].nxt
}ed[100010];
int first[100010],num_e=1;
#define f(x) first[x]
int d;
int n,m,du[100010],ans[100010],cnt;
bool v[100010];
inline void add(int u,int v)
{
	++num_e;
	u(num_e)=u;
	v(num_e)=v;
	n(num_e)=f(u);
	f(u)=num_e;
}
bool ved[100010],ving[100010];
signed main()
{
//	freopen("in.txt","r",stdin);
//	freopen("4.in","r",stdin);
//	freopen("1.out","w",stdout);

	cin>>d;
	while(d--)
	{
		ma(ed);ma(first);ma(du);ma(ans);ma(v);
		cnt=0;num_e=1;
		scanf("%d%d",&n,&m);
		int a,b;
		for(int i=1;i<=m;i++)
		{
			scanf("%d%d",&a,&b);
			add(b,a);
			du[a]++;
		}
		priority_queue<int> q;
		for(int i=1;i<=n;i++)
		if(!du[i]){v[i]=1;q.push(i);}
		if(q.empty()){puts("Impossible!");continue;}
		while(q.size())
		{
			int k=q.top();q.pop();
			ans[++cnt]=k;
			for(int i=f(k);i;i=n(i))
			if(!v[v(i)])
			{
				du[v(i)]--;
				if(!du[v(i)])
				{
					v[v(i)]=1;
					q.push(v(i));
				}
			}

		}
		if(cnt!=n){puts("Impossible!");continue;}
		for(int i=cnt;i;i--)
			printf("%d ",ans[i]);puts("");
	}
}

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<queue>
 #define ma(x) memset(x,0,sizeof(x))
 using namespace std;
 struct edge
 {
     int u,v,nxt;
     #define u(x) ed[x].u
     #define v(x) ed[x].v
     #define n(x) ed[x].nxt
 }ed[];
 int first[],num_e=;
 #define f(x) first[x]
 int d;
 int n,m,du[],ans[],cnt;
 bool v[];
 int minn[];
 bool pd=;
 inline void add(int u,int v)
 {
     ++num_e;
     u(num_e)=u;
     v(num_e)=v;
     n(num_e)=f(u);
     f(u)=num_e;
 }
 priority_queue<int,vector<int>,greater<int> >q[],temm;
 void dfs1(int x,int fa)
 {
     v[x]=;q[x].push(x);
     for(int i=f(x);i;i=n(i))
     if(v(i)!=fa)
     {
         if(!v[v(i)])dfs1(v(i),x);
         temm=q[v(i)];
         while(temm.size())
         {
             q[x].push(temm.top());
             temm.pop();
         }
     }
 }
 bool ved[],ving[];
 void dfs2(int x)
 {
 //    cout<<x<<endl;
     if(pd)return;
     ving[x]=;
 //    cout<<q[x].size()<<endl;
 //    while(!q[x].empty()&&q[x].top()==x)q[x].pop();
 //    cout<<q[x].size()<<endl;
     while(q[x].size())
     {
         while(!q[x].empty()&&q[x].top()==x)q[x].pop();
         if(q[x].empty())break;
         int k=q[x].top();q[x].pop();
 //        cout<<k<<endl;
         if(ving[k]){pd=;return;}
         if(!ved[k])dfs2(k);
     }
     ving[x]=;ved[x]=;
     ans[++cnt]=x;
 }
 signed main()
 {
 //    freopen("in.txt","r",stdin);
 //    freopen("1.in","r",stdin);
 //    freopen("1.out","w",stdout);

     cin>>d;
     while(d--)
     {
         ma(ed);ma(first);ma(du);ma(ans);ma(v);ma(minn);ma(q);ma(ving);ma(ved);
         pd=cnt=;num_e=;
         scanf("%d%d",&n,&m);
         int a,b;
         for(int i=;i<=m;i++)
         {
             scanf("%d%d",&a,&b);
             add(b,a);
             du[b]++;
         }
         for(int i=;i<=n;i++)
         if(!v[i])dfs1(i,);
         for(int i=;i<=n;i++)
         if(!ved[i])dfs2(i);
 /*        for(int i=1;i<=n;i++)
         {
             printf("#%d:\n",i);
             while(q[i].size())
             {
                 cout<<q[i].top()<<endl;
                 q[i].pop();
             }
         }*/

         if(pd){puts("Impossible!");continue;}
         for(int i=;i<=cnt;i++)
             printf("%d ",ans[i]);puts("");
     }
 }

附十万个堆代码