原題
Sorting It All Out
Time Limit: 1000MS Memory Limit: 10000K
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.
題意
依次給定一組字母的大小關係，並及時判斷它們是否能組成唯一的拓撲序列。
涉及知識及算法
題解的強大之處在於靈活利用了入度這個概念，存在解的時候，必須滿足只有一個單源點，沿着解這條路走，必然是單源點（入度爲零）後面跟着入度爲1的點， 然後刪除單源點，更新入度，重複，又是僅有一個單源點，且後面跟着入度爲1的點，最後僅剩一個單源點。有環的話，隨着單源點的刪除，最後會出現沒有單源點的情況，這個時候就可以判讀處理有環。 無序的話，無論什麼時候，出現多個單源點都說明無序。
——引自CSDN博主chchlh，附上鍊接http://blog.csdn.net/chchlh/article/details/41846509，表示感謝。
代碼
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
int n,m;
//最終判斷的標記
bool Sign;
//錄入字符串
string str;
//入度數組
int indegree[27];
//圖
int Map[27][27];
//解空間(隊列)
int q[27];
//備份入度頂點
int temp[27];
//拓撲排序
int TopoSort()
{
//記錄解空間中零入度頂點的個數
int Count=0;
//記錄任一個零入度頂點的位置
int loc;
//記錄當前圖中零入度頂點數目
int m;
//暫標記爲有序
int flag=1;
for(int i=1;i<=n;i++)
{
temp[i]=indegree[i];
}
for(int i=1;i<=n;i++)
{
m=0;
//查找零入度頂點個數
for(int j=1;j<=n;j++)
{
if(temp[j]==0)
{
m++;
//記錄一個零入度頂點位置
loc=j;
}
}
//當前圖中零入度頂點數目爲零一定說明有環
if(m==0)
{
return 0;
}
//無序，但不一定知道是否有環
if(m>1)
{
flag=-1;
}
//該零入度頂點入隊
q[Count++]=loc;
//入度置爲-1
temp[loc]=-1;
//刪除該點
for(int j=1;j<=n;j++)
{
if(Map[loc][j]==1)
{
temp[j]--;
}
}
}
return flag;
}
int main()
{
//freopen("in.txt","r",stdin);
while(scanf("%d%d",&n,&m)&&(n||m))
{
memset(Map,0,sizeof(Map));
memset(indegree,0,sizeof(indegree));
Sign=0;
for(int i=1;i<=m;i++)
{
cin>>str;
//一旦確定結果，就對後續的輸入不再操作
if(Sign)
{
continue;
}
int u=str[0]-'A'+1;
int v=str[2]-'A'+1;
Map[u][v]=1;
//入度加一
indegree[v]++;
int s=TopoSort();
//有環
if(s==0)
{
printf("Inconsistency found after %d relations.\n",i);
Sign=1;
}
//有序
else if(s==1)
{
printf("Sorted sequence determined after %d relations: ",i);
for (int j=0; j<n; j++)
{
putchar(q[j]+'A'-1); //輸出字符 putchar(ASCII)
}
printf(".\n");
Sign=1;
}
}
//無法得出結果
if(!Sign)
{
printf("Sorted sequence cannot be determined.\n");
}
}
return 0;
}
/**************************************************************
Language: C++
Result: Accepted
Time:0 ms
Memory:1508 kb
****************************************************************/代碼引自新浪博客博主康文騏，附上鍊接http://blog.sina.com.cn/s/blog_676070110100kii1.html，表示感謝。