poj3687--Labeling Balls

Labeling Balls

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 9968 Accepted: 2743

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:

1. No two balls share the same label.
2. 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

54 04 11 14 21 22 14 12 14 13 2

Sample Output

1 2 3 4-1-12 1 3 41 3 2 4

Source

POJ Founder Monthly Contest – 2008.08.31, windy7926778

要求出第几轻的是第几个球，并且要求尽可能的使输出的字典序小

这就要求如果由小到大排序的话，需要尽量让球号小的球在前，如果正序排列的话，只能保证在当前的入度为0的点是按字典序排列，可能存在如果不为0的点中存在序号小的，却被排在了后面。

采用倒序排列，由重到轻拓扑排序，保证尽量让序号大的先被拓扑，那么序号小的自然会在后面。

#include <stdio.h>#include <string.h>struct node{    int v ;    node *next ;} *head[300];int in[300] ;int p[300] , top ;void add(int u,int v){    node *q = new node ;    q->v = u ;    q->next = head[v];    head[v] = q ;}int f(int n){    int i , ans = 0 ;    top = 0 ;    while(1)    {        int u = -1 ;        for(i = n ; i >= 1 ; i--)        {            if(in[i] == 0)            {                p[top++] = i ;                in[i] = -1 ;                ans++ ;                u = i ;                break;            }        }        if(u == -1)            return -1 ;        if(ans == n)            return 1 ;        for( node *q = head[u] ; q != NULL ; q = q->next )            in[ q->v ]-- ;    }}int main(){    int t , i , j , n , m ;    scanf("%d", &t);    while(t--)    {        memset(in,0,sizeof(in));        scanf("%d %d", &n, &m);        for(i = 1 ; i <= n ; i++)            head[i] = NULL ;        for(i = 0 ; i < m ; i++)        {            int u , v ;            scanf("%d %d", &u, &v);            in[u]++ ;            add(u,v);        }        int k = f(n);        if(k == -1)            printf("-1\n");        else        {            for(i = 1 ; i <= n ; i++)            {                for(j = 0 ; j < top ; j++)                    if(p[j] == i)                        break;                if(i == n)                    printf("%d\n", n-j);                else                    printf("%d ", n-j);            }        }    }    return 0;}