nyoj 349&Poj 1094 Sorting It All Out——————【拓扑应用】

Sorting It All Out

时间限制： 3000 ms  |  内存限制：65535 KB

难度： 3

 

描述

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.

 

输入

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.

输出

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. 

样例输入

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

样例输出

Sorted sequence determined after 4 relations: ABCD.

Inconsistency found after 2 relations.

Sorted sequence cannot be determined.



题目大意：给你n个点，给你m条边代表大小关系。问你在第几条边加入后有矛盾（有环）或能确定关系，或者不能确定关系。
解题思路：首先每次加入一条边，就用floyd传递闭包，之后再判断是否形成环。如果没有环，就判断是否能确定唯一大小关系，这里有一个重要的判断条件即如果所有的结点的度等于n-1，则拓扑排序记录路径。

#include<bits/stdc++.h>

using namespace std;

int Map[50][50],indegree[50],outdegree[50];

char S_ord[50];

bool floyd(int n){

    for(int k=0;k<n;k++){   //传递闭包

        for(int i=0;i<n;i++){

            for(int j=0;j<n;j++){

                if(Map[i][k]&&Map[k][j])

                    Map[i][j]=1;

            }

        }

    }

    for(int i=0;i<n;i++)    //判断是否形成环

        if(Map[i][i])

            return 1;

    return 0;

}

bool calcu_is_ord(int n){  //计算目前是否有序

    memset(indegree,0,sizeof(indegree));

    memset(outdegree,0,sizeof(outdegree));

    for(int i=0;i<n;i++){

        for(int j=0;j<n;j++){

            if(Map[i][j]){

                indegree[j]++;

                outdegree[i]++;

            }

        }

    }

    for(int i=0;i<n;i++){

        if(indegree[i]+outdegree[i]!=n-1){  

/*如果所有结点都满足入度加出度等于结点总数减一，说明已经有序。因为如果有序，必然

会有入度为0~n-1，相应的出度为n-1~0。所以只要所有的结点度都为n-1，则说明已经有序。

*/

            return 0;

        }

    }

    return 1;

}

void topo_sort(int n){  //拓扑排序求大小顺序

    int que_[50],vis[50],top=0,cnt=0,u;

    for(int i=0;i<n;i++){

        if(indegree[i]==0){

            que_[++top]=i;

        }

    }

    memset(vis,0,sizeof(vis));

    while(top){

        u=que_[top--];

        vis[u]=1;

        S_ord[cnt++]=u+'A';

        for(int i=0;i<n;i++){

            if(!vis[i]&&Map[u][i]){

                indegree[i]--;

            }

            if(!vis[i]&&indegree[i]==0){

                que_[++top]=i;

            }

        }

    }

    S_ord[cnt++]='\0';

}

int main(){

    int n,m;

    char str[10];

    while(scanf("%d%d",&n,&m)!=EOF&&(n+m)){

        memset(Map,0,sizeof(Map));

        int flag_cir=0,flag_ord=0;  //记录在第几组关系输入时形成环或有序

        for(int i=1;i<=m;i++){

            scanf("%s",str);

            Map[str[0]-'A'][str[2]-'A']=1;

            if(flag_cir||flag_ord)

                continue;

            if(floyd(n)){ flag_cir=i;continue;}

            else if(calcu_is_ord(n)){topo_sort(n);flag_ord=i;continue;}

        }

        if(flag_cir)

            printf("Inconsistency found after %d relations.\n",flag_cir);

        else if(flag_ord){

            printf("Sorted sequence determined after %d relations: %s.\n",flag_ord,S_ord);

        }else{

            printf("Sorted sequence cannot be determined.\n");

        }

    }

    return 0;

}