Sorting It All Out
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 39602 | Accepted: 13944 |
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.
Source
分析：
有向图判环：拓扑排序
无向图判环：并查集
本题需要进行m次的拓扑排序
拓扑排序判环：拓扑之后，如果存在没有入队的点，那么该点一定是环上的点
拓扑路径的唯一性确定：队中某一时刻存在两个元素，则至少有两条不同的拓扑路径
#include<stdio.h>
#include<iostream>
#include<math.h>
#include<string.h>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<algorithm>
using namespace std;
typedef long long LL;
int mon1[]= {,,,,,,,,,,,,};
int mon2[]= {,,,,,,,,,,,,};
int dir[][]= {{,},{,-},{,},{-,}}; int getval()
{
int ret();
char c;
while((c=getchar())==' '||c=='\n'||c=='\r');
ret=c-'';
while((c=getchar())!=' '&&c!='\n'&&c!='\r')
ret=ret*+c-'';
return ret;
} #define max_v 55
int indgree[max_v];
int temp[max_v];
int G[max_v][max_v];
int tp[max_v];
int n,m;
queue<int> q;
int tpsort()
{
while(!q.empty())
q.pop();
for(int i=;i<=n;i++)
{
indgree[i]=temp[i];
if(indgree[i]==)
q.push(i);
} int c=,p;
int flag=;
while(!q.empty())
{
if(q.size()>)
flag=;
p=q.front();
q.pop();
tp[++c]=p;
for(int i=;i<=n;i++)
{
if(G[p][i])
{
indgree[i]--;
if(indgree[i]==)
q.push(i);
}
}
}
/*
拓扑完之后，存在没有入队的点，那么该点就一定是环上的
*/
if(c!=n)//存在环
return ;
else if(flag)//能拓扑但存在多条路
return ;
return -;//能拓扑且存在唯一拓扑路径
}
int main()
{
int x,y;
char c1,c2;
while(~scanf("%d %d",&n,&m))
{
if(n==&&m==)
break;
memset(G,,sizeof(G));
memset(temp,,sizeof(temp));
memset(tp,,sizeof(tp));
int flag1=,index1=;//是否有环及环的位置
int flag2=,index2=;//能否拓扑和拓扑的位置
for(int i=;i<=m;i++)
{
getchar();
scanf("%c<%c",&c1,&c2);
x=c1-'A'+;
y=c2-'A'+;
if(flag1==&&flag2==)
{
if(G[y][x])//环的一种情况
{
flag1=;
index1=i;
continue;
}
if(G[x][y]==)//预防重边
{
G[x][y]=;
temp[y]++;
}
int k=tpsort();
if(k==)//存在环
{
flag1=;
index1=i;
continue;
}else if(k==-)//存在唯一拓扑路径
{
flag2=;
index2=i;
}
}
}
if(flag1==&&flag2==)
{
printf("Sorted sequence cannot be determined.\n");
}else if(flag1)
{
printf("Inconsistency found after %d relations.\n",index1);
}else if(flag2)
{
printf("Sorted sequence determined after %d relations: ",index2);
for(int i=;i<=n;i++)
{
printf("%c",tp[i]+'A'-);
}
printf(".\n");//！！！注意还有个点...
}
}
return ;
}