poj-3687Labeling Balls(反向建图+优先队列+逆向输出)

Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12675		Accepted: 3647

Description

Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 to N 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 with b".

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 next M line each contain two integers a and b indicating
the ball labeled with a must be lighter than the one labeled with b. (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 label N. 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

这个题和逃生那个题是一类的，都是要尽量把小的往前排，所以要反向建图，如果用到邻接表的话，就还要用到优先队列

我用了两种方法

1，用到邻接表（反向拓扑+优先队列+逆向输出）

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int head[500],degree[500],result[450],m,n,t;
struct node {
int to,next;
}s[400000];
void add(int x,int y)
{
s[t].to=y;
s[t].next=head[x];
head[x]=t++;
}
int topo()
{
int i,j,p;
priority_queue<int >Q;
for(i=1;i<=m;i++)
if(degree[i]==0)
Q.push(i);
p=m;
while(!Q.empty())
{
int w=Q.top();
Q.pop();
result[w]=p--;

for(int k=head[w];k!=-1;k=s[k].next)
{
int e=s[k].to;
degree[e]--;

if(!degree[e])
Q.push(e);
}
}
if(p==0)
return 1;

return 0;
}
int main()
{
int N,a,b,i,j;
scanf("%d",&N);
while(N--)
{
t=0;
memset(degree,0,sizeof(degree));
memset(head,-1,sizeof(head));
scanf("%d%d",&m,&n);
for(i=0;i<n;i++)
{
scanf("%d%d",&a,&b);
add(b,a);
degree[a]++;
}
int qw=topo();
if(qw)
{
for(i=1;i<m;i++)
printf("%d ",result[i]);
printf("%d\n",result[i]);
}
else
printf("-1\n");
}
return 0;
}

2，反向建图，正向输出

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
int map[500][500],degree[500],m,n;
void topo()
{
int i,j,a,result[500],t=0,mark;
for(i=m;i>0;i--)
{
mark=-1;
for(j=m;j>=1;j--)
{
if(degree[j]==0)
{
mark=j;
break;
}
}
if(mark==-1)
break;
degree[mark]=-1;
result[j]=i;
for(int k=1;k<=m;k++)
{
if(map[mark][k])
degree[k]--;
}
}
if(mark==-1)
printf("-1\n");
else
{
for(i=1;i<m;i++)
printf("%d ",result[i]);
printf("%d\n",result[i]);
}
}
int main()
{
int N,i,j,a,b;
scanf("%d",&N);
while(N--)
{
memset(map,0,sizeof(map));
memset(degree,0,sizeof(degree));
scanf("%d%d",&m,&n);
for(i=0;i<n;i++)
{
scanf("%d%d",&a,&b);
if(!map[b][a])
{
map[b][a]=1;
degree[a]++;
}
}
topo();
}
return 0;
}