POJ 1094 -- Sorting It All Out (topological sorting)
题意:给你一些大写字母间的大小排序关系,判断以下3中情况:1 能唯一确定它们的排列顺序,2 所给关系是矛盾的,3 到最后也不能确定它们之间的关系。
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.
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.http://www.cnblogs.com/Jason-Damon/archive/2012/04/28/2474436.html
题意:给你一些大写字母间的大小排序关系,判断以下3中情况:1 能唯一确定它们的排列顺序,2 所给关系是矛盾的,3 到最后也不能确定它们之间的关系。
1,不存在拓扑排序,就是表明这些表达式存在矛盾
2,如果存在唯一的拓扑排序,就可以输出结果
3,如果不存在唯一的排序,即存在入度相同的点,此时表示不能确定排序关系或者存在结果矛盾(所以在不能确定排序的时候,还要判断是不是存在环,从而确定是不是存在拓扑排序)
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。
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;
}
bool adj[MAXN][MAXN]; int in_degree[MAXN]; char str[MAXN]; int n,m; int topo_sort() { int i,j,k; bool flag=true; memset(in_degree,0,sizeof(in_degree)); memset(str,'\0',sizeof(str)); for(i=1; i<=n; i++) { for(j=1; j<=n; j++) if(adj[i][j]) in_degree[j]++; //入度加一 } for(i=1; i<=n; i++) //每次产生一个字符 { k=0; for(j=1; j<=n; j++) { if(in_degree[j]==0) { if(k==0) k=j; else flag=false; //还有入度为零的节点 } } if(k==0) return 0; //没有入度为零的节点,即存在环 in_degree[k]=-1; str[i-1]=k+'A'-1; for(j=1; j<=n; j++) //k指向的节点入度都减一,即去掉A及它相关的边 { if(adj[k][j]) in_degree[j]--; } } if(flag) return 1; //没有入度为零的点,完成排序 else return 2; //排序没有完成 } int main() { int i,a,b,result; char s[4]; freopen("acm.txt","r",stdin); while(scanf("%d%d",&n,&m),m+n) { memset(adj,false,sizeof(adj)); bool h=false; for(i=1; i<=m; i++) { scanf("%s",s); a=s[0]-'A'+1; b=s[2]-'A'+1; adj[a][b]=true; if(h) continue; //必须有,因为还要继续把剩下的数据都读完 result=topo_sort(); if(result==1) { printf("Sorted sequence determined after %d relations: %s.\n",i,str); h=true; } if(result==0) { printf("Inconsistency found after %d relations.\n",i); //有换存在 h=true; } } if(!h) printf("Sorted sequence cannot be determined.\n"); } return 0; }Read full article from 1094 -- Sorting It All Out