POJ-1094 Sorting it all out (拓扑排序)

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 26070 Accepted: 9031

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.

思路：

拓扑排序（堆栈）；

需要记录的数据： 有向图（临界矩阵），  各点入度值，排序结果

先找入度为0的入栈，作为起点，放到栈顶；

再将与栈顶元素相连的点的入度减1、如果减1后有点入度为0，则入队；

类推，详见代码

代码：

#include <stdio.h>#include <string.h>#include <stack>#define N 30using namespace std;int n, m;bool map[N][N];int in[N], list[N];int toposort(){int tin[N];memcpy(tin, in, sizeof(in));stack<int>q;for(int i = 0; i < n; i ++){// 先找一个入度为1的点if(!tin[i]){q.push(i);}}int flag = 0, top, l_ct = 0;// flag 为路径唯一的标记， l_ct 为 list 数组的计数器while(!q.empty()){if(q.size() > 1)flag = 1;top = q.top();q.pop();list[l_ct ++] = top;for(int i = 0; i < n; i ++){if(map[top][i]){if(-- tin[i] == 0){// 入度为1的点入栈q.push(i);}}}}if(l_ct != n){// 不能拓扑排序，既有环（必须先判断此项）return 1;}else if(flag == 1){// 有多种排序方式return 2;}return 0;// 成功}int main(){char x, y;while(scanf("%d%d", &n, &m), n && m){int failed = 0, sure = 0;// 已失败（有环）， 已成功排序memset(map, 0, sizeof(map));memset(in, 0, sizeof(in));memset(list, 0, sizeof(list));for(int i = 1; i <= m; i ++){scanf(" %c<%c", &x, &y);if(!failed && !sure){    if(map[y - 'A'][x - 'A']){printf("Inconsistency found after %d relations.\n",i);failed = 1;continue;}if(map[x - 'A'][y - 'A'] == 0){map[x - 'A'][y - 'A'] = 1;in[y - 'A'] ++;// y点入度加1}int ans = toposort();if(ans == 0){printf("Sorted sequence determined after %d relations: ",i);for(int j = 0; j < n; j ++)printf("%c", 'A' + list[j]);printf(".\n");sure = 1;}else if(ans == 1){printf("Inconsistency found after %d relations.\n",i);failed = 1;}}}if(!failed && !sure){// 即未成功也未失败，既无法判断printf("Sorted sequence cannot be determined.\n");}}return 0;}