POJ 1094 Sorting It All Out 拓扑排序
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Sorting It All Out
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 |
看上去不就是喜闻乐见的拓扑排序?
看上去是个在线算法...实际上一点都不省时间...每次都是暴力判断是否拓扑
但在判断成环的情况时错了一个小地方,导致DEBUG了一整天...看来还是代码功力不够...敲的时候开小差...敲完以后就悲剧了...
而且我的代码复杂度奇高达到了N^4...绝对不是一个好的算法...好在数据量比较小...
本来拓扑排序邻接表的时候复杂度就是N^2了...之后我在判断是否成环的时候又用了一次N^2...结果很是悲剧...用了94MS
这代码纯属贴一下...可以用优先队列处理...复杂度能降到N+M
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <cctype>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
using namespace std;
//int max(int x,int y){return x>y?x:y;}
//int min(int x,int y){return x<y?x:y;}
char order[27];
int track[27];
int judge(int x[][27],int y[],int m)
{
int i,j,count=0,trackx=0;
int p[27][27],q[27],tra[27];
memset(tra,0,sizeof(tra));
for(i=1;i<=m;i++)
for(j=1;j<=m;j++)
p[i][j]=x[i][j];
for(i=1;i<=m;i++)
q[i]=y[i];
for(i=1;i<=26;i++)
{
if(q[i]==0 && tra[i]==0)
{
int xx=0;
tra[i]=1;
for(j=1;j<=m;j++)
{
if(p[i][j]==1)
{
p[i][j]=0;
q[j]--;
xx=1;
}
}
if(xx) i=0;
}
}
for(j=1;j<=m;j++)
if(q[j]>0 && tra[j]==0)
trackx=1;
if(i==27 && trackx==1)
return -1;//有环
for(i=1;i<=m;i++)
if(y[i]==0 && track[i]==0 )
count++;
if(count>1)
return 0;//不能确定序列
else if(count==0)
return -1; //有环
else return 1;//可以一搞
}
int topsort(int p[][27],int q[],int m)
{
int x[27][27],y[27];
int nx=0;
int i,j;
for(i=1;i<=m;i++)
for(j=1;j<=m;j++)
x[i][j]=p[i][j];
for(i=1;i<=m;i++)
y[i]=q[i];
memset(order,0,sizeof(order));
memset(track,0,sizeof(track));
for(i=1;i<=1000;i++)
{
if(nx==m)
return 1; //成功
if(judge(x,y,m)==-1) return -1;
if(judge(x,y,m)==0) return 0;
if(y[i]==0 && track[i]==0)
{
order[nx]=i+'A'-1;
nx++;
track[i]=1;
for(j=1;j<=m;j++)
{
if(x[i][j]!=0)
{
x[i][j]=0;
y[j]--;
}
}
for(j=1;j<=m;j++)
{
if(y[j]==0 && track[j]==0)
{
i=j-1;
break;
}
}
}
}
return 0;
}
int main()
{
// freopen("in.txt","r",stdin);
int m,n;
while(cin>>m>>n && m!=0 && n!=0)
{
int i,j,k;
int map[27][27];
int num[27];
memset(order,0,sizeof(order));
memset(map,0,sizeof(map));
memset(num,0,sizeof(num));
for(i = 1 ; i <= n ; i++ )
{
char a,ch,b;
cin>>a>>ch>>b;
map[b-'A'+1][a-'A'+1]=1;
num[a-'A'+1]++;
if(topsort(map,num,m)==1)
{
for(j=i+1;j<=n;j++)
cin>>a>>ch>>b;
printf("Sorted sequence determined after %d relations: ",i);
for(j=26;j>=0;j--)
if(order[j]>='A'&&order[j]<='Z')
cout<<order[j];
cout<<"."<<endl;
break;
}
if(topsort(map,num,m)==-1)
{
for(j=i+1;j<=n;j++)
cin>>a>>ch>>b;
printf("Inconsistency found after %d relations.\n",i);
break;
}
}
if(i==n+1)
printf("Sorted sequence cannot be determined.\n");
}
return 0;
}
