链接：
http://poj.org/problem?id=1094
题目：
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：
这题典型的拓扑排序，但是有点变化。
题目样例的三种输出分别是：
1. 在第x个关系中可以唯一的确定排序，并输出。
2. 在第x个关系中发现了有回环（Inconsisitency矛盾）
3.全部关系都没有发现上面两种情况，输出第3种.
那么对于给定的m个关系，一个个的读进去，每读进去一次就要进行一次拓扑排序，如果发现情况1和情况2，那么就不用再考虑后面的那些关系了，但是还要继续读完后面的关系（但不处理）。如果读完了所有关系，还没有出现情况1和情况2，那么就输出情况3.
拓扑排序有两种方法，一种是算法导论上的，一种是用贪心的思想，这题用贪心的思想做更好。
贪心的做法：
1. 找到所有入度为0的点， 加入队列Q
2.取出队列Q的一个点，把以这个点为起点，所有它的终点的入度都减1. 如果这个过程中发现经过减1后入度变为0的，把这个点加入队列Q。
3.重复步骤2，直到Q为空。
这个过程中，如果同时有多个点的入度为0，说明不能唯一确定关系。
如果结束之后，所得到的经过排序的点少于点的总数，那么说明有回环。
题目还需要注意的一点：如果边(u,v)之前已经输入过了，那么之后这条边都不再加入。
代码：
#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; // 如果已经判断了有回环或者可唯一排序，不处理但是要继续读
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];
}
// 拷贝一个副本，等下用来还原in数组
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;
}
—— 生命的意义，在于赋予它意义。
原创 http://blog.csdn.net/shuangde800 ， By D_Double (转载请标明)
