Description
John has n tasks to do. Unfortunately, the tasks are not independent and the execution of one task is only possible if other tasks have already been executed
Input
The input will consist of several instances of the problem. Each instance begins with a line containing two integers, 1 <= n <= 100 and m. n is the number of tasks (numbered from 1 to n) and m is the number of direct precedence relations between tasks. After this, there will be m lines with two integers i and j, representing the fact that task i must be executed before task j. An instance with n = m = 0 will finish the input.
Output
For each instance, print a line with n integers representing the tasks in a possible order of execution.
Sample Input
5 4
1 2
2 3
1 3
1 5
0 0
Sample Output
1 4 2 5 3
理解
拓?fù)渑判虻囊粋€(gè)比較典型的題目.題目大意是:每一項(xiàng)任務(wù)都有一個(gè)編號(hào),輸入的兩個(gè)編號(hào)表示只有做過前邊的編號(hào)任務(wù)才能接著去做后邊的編號(hào)任務(wù),讓我們輸出做任務(wù)的順序.
代碼部分
#include<iostream>
#include<cstring>
using namespace std;
int map[101][101],inde[101];
void topu(int a[101][101],int d[101],int n)
{
int f=0;
for(int i=0;i<n;i++)
{
for(int j=1;j<=n;j++)
{
if(d[j]==0)
{
d[j]--;
if(f==1)
{cout<<" "<<j;}
if(f==0)
{
cout<<j;
f=1;
}
for(int k=1;k<=n;k++)
{
if(a[j][k]==1)
{
d[k]--;
}
}
}
}
}
}
int main()
{
int m,n;
while(cin>>m>>n)
{
if(m==0&&n==0)
return 0;
int a,b;
memset(map,0,sizeof(map));
memset(inde,0,sizeof(inde));
for(int i=1;i<=n;i++)
{
cin>>a>>b;
if(!map[a][b])
{
map[a][b]=1;
inde[b]++;
}
}
topu(map,inde,m);
cout<<endl;
}
return 0;
}
