| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 9028 | Accepted: 2444 |
Description
Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 toN in such a way that:
- No two balls share the same label.
- The labeling satisfies several constrains like "The ball labeled with a is lighter than the one labeled withb".
Can you help windy to find a solution?
Input
The first line of input is the number of test case. The first line of each test case contains two integers,N (1 ≤N ≤ 200) and M (0 ≤ M ≤ 40,000). The nextM line each contain two integersa and b indicating the ball labeled witha must be lighter than the one labeled withb. (1 ≤ a, b ≤N) There is a blank line before each test case.
Output
For each test case output on a single line the balls' weights from label 1 to labelN. If several solutions exist, you should output the one with the smallest weight for label 1, then with the smallest weight for label 2, then with the smallest weight for label 3 and so on... If no solution exists, output -1 instead.
Sample Input
5 4 0 4 1
1 1 4 2
1 2
2 1 4 1
2 1 4 1
3 2
Sample Output
1 2 3 4
-1
-1
2 1 3 4
1 3 2 4
#include<iostream>
#include<stdio.h>
#include<queue>
using namespace std;
typedef struct v{ int vex;
v *next; }V;
V*p;
typedef struct h{ int indegree;
v *next ;
}H;
int n,result[300],tot,ans[300];
priority_queue<int ,vector<int>,less<int> > q;
H team[50000];
bool visit[305][305];
void topsort()
{
int i,ci,sum,a,b;
ci=1;
for(i=1;i<=n;i++)
{ if(team[i].indegree==0)
q.push(i); }
sum=n; while(!q.empty())
{
sum--;
a=q.top();
q.pop();
result[ci++]=a;
for(p=team[a].next;p!=0;p=p->next)
{ b=p->vex;
if(--team[b].indegree==0)
q.push(b);
} }
// printf("%d\n",sum);
if(sum>0)
{
printf("-1\n");
}
else
{
int nn=n;
for(i=1;i<=n;i++)
{ ans[result[i]]=nn--;//这是排序
}
for(i=1;i<n;i++) printf("%d ",ans[i]);//这是重量
printf("%d\n",ans[i]); } }
int main()
{ int i,m,a,b,T,j; scanf("%d",&T);
while(T--)
{
while(!q.empty())
{ q.pop();
} scanf("%d%d",&n,&m);
for(i=0;i<=n;i++)
for(j=0;j<=n;j++)
{ visit[i][j]=false;
} for(i=0;i<=n;i++)
{
team[i].indegree=0;
team[i].next=NULL;
result[i]=0;
}
while(m--)
{ scanf("%d%d",&b,&a);//反向建表
if(visit[a][b])
continue;
visit[a][b]=true;
team[b].indegree++;
p=new V;
p->vex=b;
p->next=team[a].next;
team[a].next=p; } topsort(); } return 0;
}