标签：ems poj data output lin case ever bool integer
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 10178 | Accepted: 2815 |
Description
Windy has N balls of distinct weights from 1 unit toN units. Now he tries to label them with 1 toN in such a way that:
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
在主要的拓扑排序的基础上又添加了一个要求：编号最小的节点要尽量排在前面；在满足上一个条件的基础上，编号第二小的节点要尽量排在前面；
在满足前两个条件的基础上。编号第三小的节点要尽量排在前面……依此类推。点击打开链接又是看结题报告。。。
。哎。。
。。
太弱了
第一百篇。。留念。。。
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;
const int M = 250 ;
int t, n, m;
int outint[M];
int out[M];
int in[M];
int cut;
vector<int>amap[M];
int flag;
bool toposort()
{
cut = 0;
priority_queue<int>que;
for(int i=1; i<=n; i++)
if( !in[i] )
que.push(i);
while( !que.empty() )
{
int u = que.top();
que.pop();
outint[ cut++] = u;
for( int i=0; i<amap[u].size(); i++ )
{
int v = amap[u][i];
if( --in[v]==0 )
que.push(v);
}
}
if( cut<n )
return false;
else
return true;
}
int main()
{
scanf( "%d", &t );
while( t-- )
{
memset( in, 0, sizeof(in) );
scanf( "%d%d", &n, &m );
for( int i=1; i<=n; i++ )
amap[i].clear();
for( int i=1; i<=m; i++ )
{
int a, b;
scanf( "%d%d", &a, &b );
amap[b].push_back(a);
in[a]++;
}
if( !toposort() )
printf("-1\n");
else
{
for( int i=0; i<n; i++ )
out[ outint[i] ] = n-i;
for(int i=1; i<=n; i++)
{
if(i<n)
printf( "%d ", out[i] );
else
printf( "%d\n", out[i] );
}
}
}
return 0;
}
POJ 3687：Labeling Balls（优先队列+拓扑排序）
标签：ems poj data output lin case ever bool integer
原文地址：http://www.cnblogs.com/mthoutai/p/7258459.html
