題目：
Sorting It All Out
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 21532 Accepted: 7403
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.
分析：
這題典型的拓撲排序，但是有點變化。
題目樣例的三種輸出分別是：
1. 在第x個關系中可以唯一的確定排序，並輸出。
2. 在第x個關系中發現了有回環（Inconsisitency矛盾）
3.全部關系都沒有發現上面兩種情況，輸出第3種.
那麼對於給定的m個關系，一個個的讀進去，每讀進去一次就要進行一次拓撲排序，如果發現情況1和情況2，那麼就不用再考慮後面的那些關系了，但是還要繼續讀完後面的關系（但不處理）。如果讀完了所有關系，還沒有出現情況1和情況2，那麼就輸出情況3.
拓撲排序有兩種方法，一種是算法導論上的，一種是用貪心的思想，這題用貪心的思想做更好。
貪心的做法：
1. 找到所有入度為0的點， 加入隊列Q
2.取出隊列Q的一個點，把以這個點為起點，所有它的終點的入度都減1. 如果這個過程中發現經過減1後入度變為0的，把這個點加入隊列Q。
3.重復步驟2，直到Q為空。
這個過程中，如果同時有多個點的入度為0，說明不能唯一確定關系。
如果結束之後，所得到的經過排序的點少於點的總數，那麼說明有回環。
題目還需要注意的一點：如果邊(u,v)之前已經輸入過了，那麼之後這條邊都不再加入。
代碼：
[cpp]
#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;
const int N = 105;
int n,m,in[N],temp[N],Sort[N],t,pos, num;
char X, O, Y;
vector<int>G[N];
queue<int>q;
void init(){
memset(in, 0, sizeof(in));
for(int i=0; i<=n; ++i){
G[i].clear();
}
}
inline bool find(int u,int v){
for(int i=0; i<G[u].size(); ++i)
if(G[u][i]==v)return true;
return false;
}
int topoSort(){
while(!q.empty())q.pop();
for(int i=0; i<n; ++i)if(in[i]==0){
q.push(i);
}
pos=0;
bool unSure=false;
while(!q.empty()){
if(q.size()>1) unSure=true;
int t=q.front();
q.pop();
Sort[pos++]=t;
for(int i=0; i<G[t].size(); ++i){
if(--in[G[t][i]]==0)
q.push(G[t][i]);
}
}
if(pos<n) return 1;
if(unSure) return 2;
return 3;
}
int main(){
int x,y,i,flag,ok,stop;
while(~scanf("%d%d%*c",&n,&m)){
if(!n||!m)break;
init();
flag=2;
ok=false;
for(i=1; i<=m; ++i){
scanf("%c%c%c%*c", &X,&O,&Y);
if(ok) continue; // 如果已經判斷了有回環或者可唯一排序，不處理但是要繼續讀
x=X-'A', y=Y-'A';
if(O=='<'&&!find(y,x)){
G[y].push_back(x);
++in[x];
}
else if(O=='>'&&!find(x,y)){
G[x].push_back(y);
++in[y];
}
// 拷貝一個副本，等下用來還原in數組
memcpy(temp, in, sizeof(in));
flag=topoSort();
memcpy(in, temp, sizeof(temp));
if(flag!=2){
stop=i;
ok=true;
}
}
if(flag==3){
printf("Sorted sequence determined after %d relations: ", stop);
for(int i=pos-1; i>=0; --i)
printf("%c",Sort[i]+'A');
printf(".\n");
}
else if(flag==1){
printf("Inconsistency found after %d relations.\n",stop);
}
else{
printf("Sorted sequence cannot be determined.\n");
}
}
return 0;
}
