POJ3678 Labeling Balls 拓扑排序

　　题目有很多坑点，首先要求编号越小的节点在结果中拍在越前面，其次是输出节点所在的位置，而不是节点的编号！！

　　如果正向建图，那么每次去掉入度为零的节点都要考虑其能到达的所有节点中的最小的，如果有最小的还要比较次小的。。。做起来很麻烦。可以考虑反向建图，那么每次选择入度为零且节点编号最大的节点，那么编号越大，越靠后，编号小的越靠前。

  1 //STATUS:C++_AC_63MS_416KB
  2 #include <functional>
  3 #include <algorithm>
  4 #include <iostream>
  5 //#include <ext/rope>
  6 #include <fstream>
  7 #include <sstream>
  8 #include <iomanip>
  9 #include <numeric>
 10 #include <cstring>
 11 #include <cassert>
 12 #include <cstdio>
 13 #include <string>
 14 #include <vector>
 15 #include <bitset>
 16 #include <queue>
 17 #include <stack>
 18 #include <cmath>
 19 #include <ctime>
 20 #include <list>
 21 #include <set>
 22 #include <map>
 23 using namespace std;
 24 //using namespace __gnu_cxx;
 25 //define
 26 #define pii pair<int,int>
 27 #define mem(a,b) memset(a,b,sizeof(a))
 28 #define lson l,mid,rt<<1
 29 #define rson mid+1,r,rt<<1|1
 30 #define PI acos(-1.0)
 31 //typedef
 32 typedef long long LL;
 33 typedef unsigned long long ULL;
 34 //const
 35 const int N=210;
 36 const int INF=0x3f3f3f3f;
 37 const int MOD=100000,STA=8000010;
 38 const LL LNF=1LL<<60;
 39 const double EPS=1e-8;
 40 const double OO=1e15;
 41 const int dx[4]={-1,0,1,0};
 42 const int dy[4]={0,1,0,-1};
 43 const int day[13]={0,31,28,31,30,31,30,31,31,30,31,30,31};
 44 //Daily Use ...
 45 inline int sign(double x){return (x>EPS)-(x<-EPS);}
 46 template<class T> T gcd(T a,T b){return b?gcd(b,a%b):a;}
 47 template<class T> T lcm(T a,T b){return a/gcd(a,b)*b;}
 48 template<class T> inline T lcm(T a,T b,T d){return a/d*b;}
 49 template<class T> inline T Min(T a,T b){return a<b?a:b;}
 50 template<class T> inline T Max(T a,T b){return a>b?a:b;}
 51 template<class T> inline T Min(T a,T b,T c){return min(min(a, b),c);}
 52 template<class T> inline T Max(T a,T b,T c){return max(max(a, b),c);}
 53 template<class T> inline T Min(T a,T b,T c,T d){return min(min(a, b),min(c,d));}
 54 template<class T> inline T Max(T a,T b,T c,T d){return max(max(a, b),max(c,d));}
 55 //End
 56 
 57 struct Edge{
 58     int u,v;
 59 }e[N*N];
 60 int first[N],next[N*N],vis[N],out[N],ans[N];
 61 int T,n,m,mt;
 62 
 63 void adde(int a,int b)
 64 {
 65     e[mt].u=a;e[mt].v=b;
 66     next[mt]=first[a],first[a]=mt++;
 67 }
 68 
 69 int topo()
 70 {
 71     int i,j,k=n,ok;
 72     mem(out,0);
 73     while(k--){
 74         mem(vis,0);
 75         for(i=1;i<=n;i++){
 76             for(j=first[i];j!=-1;j=next[j]){
 77                 vis[e[j].v]=1;
 78             }
 79         }
 80         ok=0;
 81         for(i=n;i>=1;i--)
 82             if(!vis[i] && !out[i]){ok=1;break;}
 83         if(ok==0)return 0;
 84         ans[k]=i;
 85         out[i]=1;
 86         first[i]=-1;
 87     }
 88     return 1;
 89 }
 90 
 91 int main()
 92 {
 93  //   freopen("in.txt","r",stdin);
 94     int i,a,b;
 95     scanf("%d",&T);
 96     while(T--)
 97     {
 98         scanf("%d%d",&n,&m);
 99         mem(first,-1);mt=0;
100         while(m--){
101             scanf("%d%d",&a,&b);
102             adde(b,a);
103         }
104 
105         if(topo()){
106             for(i=0;i<n;i++)vis[ans[i]]=i+1;
107             printf("%d",vis[1]);
108             for(i=2;i<=n;i++)
109                 printf(" %d",vis[i]);
110         }
111         else printf("-1");
112         putchar('\n');
113     }
114     return 0;
115 }