【POJ 1094】Sorting It All Out （邻接表）

n 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.

有如下三种情况

如果可以确定唯一的排序顺序，输出顺序

有回环

顺序不唯一

这道题稍微有点坑人，要输出经过多少步骤可以得出结论，所以每次输入之后都要进行拓扑排序进行判断，还要对输入的关系进行标记，重复输入时不作处理

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<cstdio>
#include<queue>
#include<vector>
#include<list>
using namespace std;
const int MX = 30;
int n, m, tot, s, ind, IN[MX], head[MX], rec[MX];
int S[MX], flag, vis[MX];
char ch[MX];
struct Edge{
char v;
int nxt;
}e[10010];
void add(int u, int v){
e[tot].v = v;
e[tot].nxt = head[u];
head[u] = tot++;
IN[v]++;
}
void topo_sort(){
queue<int> q;
memset(S, 0, sizeof(S));
for(int i = 0; i < n; i++)
rec[i] = IN[i];
for(int i = 0; i < n; i++)
if(vis[i] && rec[i] == 0)
q.push(i);
ind = 0;
int temp = 0;
while(!q.empty())
{
if(q.size() > 1)
temp = 1;
int v = q.front();
S[ind++] = v;
q.pop();
for(int i = head[v]; ~i; i = e[i].nxt)
{
int u = e[i].v;
rec[u]--;
if(rec[u] == 0)
{
q.push(u);
}
}
}
if(!temp && ind == n)
flag = 2;
if(ind == s && s < n)
flag = 0;
if(ind < s)
flag = 1;
}
int main()
{
char c[5];
int x, y;
while((~scanf("%d%d", &n, &m)) && (n || m))
{
memset(IN, 0, sizeof(IN));
memset(head, -1, sizeof(head));
memset(vis, 0, sizeof(vis));
s = 0;
tot = 0;
flag = 0;
for(int i = 1; i <= m; i++)
{
scanf("%s", c);
if(flag)
continue;
x = c[0] - 'A';
y = c[2] - 'A';
if(!vis[x])
{
vis[x] = 1;
s++;
}
if(!vis[y])
{
vis[y] = 1;

b261
s++;
}
add(x, y);
topo_sort();
if(flag == 2)
{
cout<<"Sorted sequence determined after "<<i<<" relations: ";
for(int i = 0; i < n; i++)
cout<<char(S[i] + 'A');
cout<<"."<<endl;
}
if(flag == 1)
{
cout<<"Inconsistency found after "<<i<<" relations."<<endl;
}
}
if(flag == 0)
cout<<"Sorted sequence cannot be determined."<<endl;
}
return 0;
}

这有一份拓扑模板，但是并不适用于本题目，不过还是会有一定的启发作用。邻接表的实现可以减少空间的浪费

#include<bits/stdc++.h>
using namespace std;
const int MX = 1e5 +5;
struct Edge{
int v,nxt;
}E[MX*2];
int head[MX],tot,IN[MX];
void init(){
memset(head,-1,sizeof(head));
tot=0;
memset(IN,0,sizeof(IN));
}
void add(int u,int v){
E[tot].v=v;
E[tot].nxt=head[u];
head[u]=tot++;
IN[v]++;
}
int vec[MX],sz;
bool top_sort(int n){
sz=0;
queue<int>q;
for(int i=1;i<=n;i++) if(IN[i]<=1) q.push(i);
while(!q.empty()){
int u=q.front();q.pop();
vec[++sz]=u;
for(int i=head[u];~i;i=E[i].nxt){
int v=E[i].v;
IN[v]--;
if(IN[v]<=1) q.push(v);
}
}
if(sz!=n) return 0;
return 1;
}
int main(){
int n,m;
while(~scanf("%d%d",&n,&m)){
init();
for(int i=1;i<=m;i++) {
int u,v;
scanf("%d%d",&u,&v);
add(u,v);add(v,u);
}
if(top_sort(n)) printf("YES\n");
else printf("NO\n");
}
return 0;
}