nyoj 349&Poj 1094 Sorting It All Out

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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;
}

　　

nyoj 349&Poj 1094 Sorting It All Out

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