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  • 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++]='';
    }
    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.
    ",flag_cir);
            else if(flag_ord){
                printf("Sorted sequence determined after %d relations: %s.
    ",flag_ord,S_ord);
            }else{
                printf("Sorted sequence cannot be determined.
    ");
            }
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/chengsheng/p/4462267.html
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