POJ-1094 Sorting it all out (拓扑排序)

Sorting It All Out

Time Limit: 1000MS		Memory Limit: 10000K
Total 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 6
A<B
A<C
B<C
C<D
B<D
A<B
3 2
A<B
B<A
26 1
A<Z
0 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 30

using namespace std;

int n, m;
bool map

;
int in
, list
;

int toposort()
{
int tin
;
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;
}