POJ 1094 -- Sorting It All Out （拓扑排序）

Sorting It All Out
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice POJ
1094

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.

题意：根据给出的关系式判断大小关系，当出现环时输出在第几个关系式出现环，

当可排序时输出在第几个关系式可排序并输出排序顺序。

当所有关系式输入后无法排序则输出无法确定排序。

思路：因为要输出第几个关系式，所以每输入一个关系式都要对图进行拓扑排序，

若发现环则输出含有环，若可排序则输出可排序，重点在于 当出现多个0

前驱节点时，应将所有0前驱标记，继续对图排序，看看是否含有环！！

输出遵照以下规则：优先级：输出环，输出完整顺序，输出无法排序。

代码如下：

法1：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
#define s 120
#define inf 0x3f3f3f3f
int n, m, indegree[s], vis[s], tmp[s];
char c[4], ans[s];
vector<vector<int>> v;//相当于二维数组
int toposort(int obj)
{
int num, flag = 1, k, p = 0, i;
for (i = 0; i < n; i++)  tmp[i] = indegree[i];//临时存放入度数组,以免影响下次结果
while (obj--)
{
num = 0;//每次重新计数
for (i = 0; i < n; i++){
if (tmp[i] == 0){
k = i, num++;
}
}
if (num > 0){
if (num > 1)  flag = 0;
else{
ans[p++] = k + 'A';
ans[p] = '\0';
}
for (i = 0; i < v[k].size(); i++){
tmp[v[k][i]]--;
}
tmp[k]--;
}
else
return -1;//数据矛盾
}
if (flag) return p;//返回能排序的位数
return 0;//排序不确定
}
int main()
{
#ifdef OFFLINE
freopen("t.txt", "r", stdin);
#endif
int i, j, k;
while (~scanf("%d %d", &n, &m) && n&&m)
{
v.clear();   v.resize(n);
memset(vis, 0, sizeof(vis));
memset(indegree, 0, sizeof(indegree));
int obj = 0, det = 0;
for (i = 1; i <= m; i++){
scanf("%s", c);
indegree[c[2] - 'A']++;//入度
v[c[0] - 'A'].push_back(c[2] - 'A');//建边
if (!vis[c[0] - 'A']){
vis[c[0] - 'A'] = 1;
obj++;//对象数
}
if (!vis[c[2] - 'A']){
vis[c[2] - 'A'] = 1;
obj++;
}
if (det == 0)//排序不确定才继续
{
int res = toposort(obj);
if (res == -1){
k = i, det = -1;//k记录数据组数
}
else if (res == n){
k = i, det = 1;
}
}
}
if (det == -1)
printf("Inconsistency found after %d relations.\n", k);
else if (det == 0)
printf("Sorted sequence cannot be determined.\n");
else
printf("Sorted sequence determined after %d relations: %s.\n", k, ans);
}
return 0;
}

法二（与法一差不多，好理解点）

#include <iostream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <vector>
#include <algorithm>
using namespace std;
#define s 120
#define inf 0x3f3f3f3f
int n, m, indegree[s], vis[s], tmp[s], rel[s][s];
char c[4], ans[s];
int toposort()
{
int i, j, zero, k, flag = 1;//sure
memcpy(tmp, indegree, sizeof(tmp));//将长度为数组tmp长度的indegree数组数据复制到tmp数组
memset(ans, '\0', sizeof(ans));
for (i = 0; i < n; i++)
{
zero = 0;//入度为0的点的个数
for (j = 0; j < n; j++){
if (tmp[j] == 0){
k = j;//下标
zero++;
}
}
if (zero == 0) return -1;//有环
if (zero>1)  flag = 0;//无序（还不能确定是否有环, 所以继续）
else    ans[i] = 'A' + k;
tmp[k]--;
for (j = 0; j < n; j++){
if (rel[k][j])//由此点出发能直达的点入度减 1
tmp[j]--;
}
}
return flag;
}
int main()
{
#ifdef OFFLINE
freopen("t.txt", "r", stdin);
#endif
int i, j, k;
while (~scanf("%d %d", &n, &m) && n + m)
{
bool sure = false;
memset(rel, 0, sizeof(rel));
memset(indegree, 0, sizeof(indegree));
for (i = 1; i <= m; i++)
{
scanf("%s", c);
if (sure) continue;
if (!rel[c[0] - 'A'][c[2] - 'A']){//判重,避免加入重复边
rel[c[0] - 'A'][c[2] - 'A'] = 1;
indegree[c[2] - 'A']++;
}
int res = toposort();
if (res == 1){
sure = true;
printf("Sorted sequence determined after %d relations: %s.\n", i, ans);
}
else if (res == -1){
sure = true;
printf("Inconsistency found after %d relations.\n", i);
}
}
if (!sure)
printf("Sorted sequence cannot be determined.\n");
}
return 0;
}