poj 1094 Sorting It All Out 【拓扑排序】

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 29404 Accepted: 10184

Description

An ascending sorted sequence of distinct values is one in which some form of a less-than operator is used to order the elements from smallest to largest. For example, the sorted sequence A, B, C, D implies that A < B, B < C and C < D. in this problem, we will give you a set of relations of the form A < B and ask you to determine whether a sorted order has been specified or not.

Input

Input consists of multiple problem instances. Each instance starts with a line containing two positive integers n and m. the first value indicated the number of objects to sort, where 2 <= n <= 26. The objects to be sorted will be the first n characters of the uppercase alphabet. The second value m indicates the number of relations of the form A < B which will be given in this problem instance. Next will be m lines, each containing one such relation consisting of three characters: an uppercase letter, the character "<" and a second uppercase letter. No letter will be outside the range of the first n letters of the alphabet. Values of n = m = 0 indicate end of input.

Output

For each problem instance, output consists of one line. This line should be one of the following three:

Sorted sequence determined after xxx relations: yyy...y.
Sorted sequence cannot be determined.
Inconsistency found after xxx relations.

where xxx is the number of relations processed at the time either a sorted sequence is determined or an inconsistency is found, whichever comes first, and yyy...y is the sorted, ascending sequence.

Sample Input

4 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 0

Sample Output

Sorted sequence determined after 4 relations: ABCD.Inconsistency found after 2 relations.Sorted sequence cannot be determined.

费解。。。 得到拓扑序列 或者 成环没有拓扑序列 的优先级高于 无序。 题目意思是：只要在 当前输入的关系 可以得到拓扑序列 或者 成环 就直接输出answer 。

站在别人的肩膀上ac的。 第一道拓扑序列：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int map[27][27];int in[27];//记录每个点的入度int mark[27];//存储每次查询时各点的入度 int topo[27];//存储序列值 int n, m;void trans()//传递每个节点的入度 {    for(int i = 1; i <= n; i++)    mark[i] = in[i];}int toposort(){    trans();    int i, j;    int t = 0;//统计当前序列个数     int num;//统计当前入度为0的点的个数     int next;//数组 下一并入点     int judge = 1;    for(i = 1; i <= n; i++)    {        num = 0;        for(j = 1; j <= n; j++)        {            if(!mark[j])//入度为0             {                num++;                next = j;            }        }        if(!num) //有环 查询失败        return 0;         if(num > 1)//当前无序 这里只能标记不能返回 因为有可能在后面查询中会成环 若成环直接输出答案         judge = -1;        //return -1;        topo[t++] = next;//并入数组         //printf("%d\n", next);        mark[next] = -1;//不再查询该点         for(j = 1; j <= n; j++)//更新操作 与并入点有关系的点入度全部减一          {            if(map[next][j])            mark[j]--;        }    }    return judge;     //return 1;} void init(){    for(int i = 1; i <= n; i++)    {        in[i] = 0;        for(int j = 1; j <= n; j++)        {            map[i][j] = 0;        }    }}int main(){    int i, j;    int x, y;    int sign;//标记是否查询出结果     int exist;    char str[10];    while(scanf("%d%d", &n, &m),n||m)    {        init();//初始化         sign = 0;//若查询出结果标记为1         for(i = 1; i <= m; i++)        {            scanf("%s", str);            if(sign) continue;//已有结果 下面的不用处理            x = str[0] - 'A' + 1;            y = str[2] - 'A' + 1;            map[x][y] = 1;            in[y]++;//记录入度            exist = toposort();             if(exist == 1)//查询成功             {                sign = 1;//已经查询出结果                 printf("Sorted sequence determined after %d relations: ", i);                for(j = 0; j < n; j++)                {                    printf("%c", 'A'+topo[j]-1);                }                printf(".\n");            }            else if(!exist) //矛盾 ->查询失败             {                sign = 1;//已经有结果                printf("Inconsistency found after %d relations.\n", i);             }         }        if(!sign)        printf("Sorted sequence cannot be determined.\n");     }    return 0;}