poj 1094 Sorting It All Out(拓扑排序)

Sorting It All Out

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 27572 Accepted: 9546

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 6A<BA<CB<CC<DB<DA<B3 2A<BB<A26 1A<Z0 0

Sample Output

Sorted sequence determined after 4 relations: ABCD.Inconsistency found after 2 relations.Sorted sequence cannot be determined.

Source

East Central North America 2001

        当然要建图。。一开始准备用dfs跑下去，dfs的思路是这样的：每加入一条边，若已确定有唯一的顺序或存在矛盾解就不管他，否则就对所有的点dfs，每访问过一个点就标记下来，若后来访问的点能再次访问已标记的点（存在环）则说明此序列是矛盾的，最终的结果也一定是矛盾的；若能够找到找到一个解，且此解的长度刚好是n,则说明结果一定是存在唯一解。关键是那种存在多解的情况，找到一个解，此解的长度 <n ，不能说明当前已有的边为多解，还可能是矛盾！例如  存在图： 1->2,3->4,4->5,5->3.以1为起点开始dfs  无论如何也检测不到3,4,5的那个环，这就决定了我们必须要对图中当前入度为0的节点全都dfs一次。另外还有一些细节问题，如：进行下一次dfs前，就要判断当前点所能到的所有点是否已标记，是则返回矛盾的解。。。顿时崩溃了。早已写了好长。难道天要亡我~~。只能回到传统的拓扑排序解法。

【思路】

每加入一条边，就拓扑排序一次，拓扑排序的时候要删除入度，排完序要恢复，关键是判断是否有多解，代码中说明

【代码】

#include<cstdio>#include<cstring>#include<iostream>using namespace std;int n;int R[30][30];int rudu[30],trudu[30];int Q[30];int topo[30],ntopo;char ansr[30];void copytopo(){    for(int i=0;i<n;i++)        ansr[i]=topo[i]+'A';    ansr[n]=0;}void copyrudu(){    for(int i=0;i<n;i++)        trudu[i]=rudu[i];}int toposort(){    ntopo=0;    int front=0,rear=0;    for(int i=0;i<n;i++)        if(rudu[i]==0)            Q[rear++]=i;    copyrudu();    if(rear==0)return 2;    int p2=rear;//表示可能为多解    while(front<rear)    {        int temp=Q[front++];        topo[ntopo++]=temp;        int cnt=0;        for(int i=0;i<n;i++)        {            if(R[temp][i]&&!(--trudu[i]))            {                Q[rear++]=i;                cnt++;            }        }        if(cnt>1)p2=2;//一下子有多个入度为0的点入队，即表明可能为多解    }    for(int i=0;i<n;i++)        if(trudu[i])            return 2;//若存在入度还不为0的点，就一定有环！    if(p2>1)return 0;    if(ntopo==n)    {        copytopo();//有唯一解就赋给终解        return 1;    }    return 0;}int main(){    int m;    while(scanf("%d%d",&n,&m)!=EOF)    {        if(n==0&&m==0)break;        memset(R,0,sizeof(R));        memset(rudu,0,sizeof(rudu));        int flag=0;//1表示有唯一解，2表示有矛盾的解；        int ans;        for(int i=1;i<=m;i++)        {            char S[10];            scanf("%s",S);            if(flag)continue;//唯一解和矛盾解都不管            ans=i;            if(!R[S[0]-'A'][S[2]-'A'])//有重边                rudu[S[2]-'A']++;            R[S[0]-'A'][S[2]-'A']=1;            flag=toposort();        }        if(!flag)            printf("Sorted sequence cannot be determined.\n");        else        {            if(flag==1)                printf("Sorted sequence determined after %d relations: %s.\n",ans,ansr);            else                printf("Inconsistency found after %d relations.\n",ans);        }    }}

最后还因为输出WA两次。。OMG