ICode9

精准搜索请尝试: 精确搜索
首页 > 其他分享> 文章详细

POJ1094 Sorting It All Out 拓扑排序

2019-09-03 10:01:39  阅读:300  来源: 互联网

标签:Sorting POJ1094 sequence int relations 次序 sorted include Out


思路借鉴:https://www.cnblogs.com/yueshuqiao/archive/2011/08/16/2140485.html
题意:A~Z中n个字母,给你m对先后关系,让你判断它们的顺序情况。
思路:有以下三种情况:
1、有环,说明给定次序矛盾。
2、能确定唯一次序。
3、无法确定唯一次序。
其中1、2可以不用完m对先后关系,而3必须用完m对先后关系才能确定。
因为m对关系是逐一加入的,所以每当加入一对关系就要进行一次拓扑排序(数据量不大,可以这么造)。
只要出现环,或者已经排好序,之后的数据就不再处理(尽管排好序后出现环也不管)。
若所有数据处理完后,还是没有确定次序,就输出不能确定次序。
当然这里有很多剪枝,比如当输入的次序之前出现相反的次序时,一定有环等等。但这里数据量不大,剪枝也没必要。
下面是AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define ll long long
#define inf 0x3f3f3f3f
#define cl(a,b) memset(a,b,sizeof(a))
using namespace std;
int n,m;
int in[30],map[30][30],ans[30];
int topo(){//三种返回的情况
			//这里cnt要初始化!!! 
	int num,pos,cnt=0;//统计入度为0的个数 和入度为0的位置 和入队的个数 
	int temp[30];
	for(int i=1;i<=n;i++){//复制一份入度 
		temp[i]=in[i];
	}
	int flag=1;//标记是否可确定唯一顺序 
	for(int i=1;i<=n;i++){ 
		num=0;//因为每次扫都要统计一下入度为0的次数
		for(int j=1;j<=n;j++){
			if(temp[j]==0){
				num++;
				pos=j; 
			}
		}
		if(num==0) return 0;//如果没有入度为0的 说明有环
		if(num>=2) flag=-1;//同时出现两个头头,不能判定有序 又怕后面出现
		ans[++cnt]=pos;
		for(int j=1;j<=n;j++){
			if(map[pos][j]){
				temp[j]--;
			}
		}
		temp[pos]--;//入队的这个度数就不管了 
	}
	return flag;//只要有一次让flag=-1 那就一定是-1 那后面的处理只是为了判断有环 
}
int main(){
	while(scanf("%d%d",&n,&m)!=EOF){
		if(n==0&&m==0) break;
		cl(in,0);
		cl(map,0);
		cl(ans,0);
		char s[5];
		int flag=-1;
		for(int i=1;i<=m;i++){//对于每次增加的边
			scanf("%s",s);
			if(flag==0||flag==1) continue;
			int u=s[0]-'A'+1,v=s[2]-'A'+1;
			in[v]++;
			map[u][v]=1;
			flag=topo();
			if(flag==0){//有环 
				printf("Inconsistency found after %d relations.\n",i); 
			}if(flag==1){//能排序 
				printf("Sorted sequence determined after %d relations: ",i);
				for(int i=1;i<=n;i++){
					printf("%c",(char)(ans[i]+'A'-1));
				}
				printf(".\n");
			}
		}
		if(flag==-1){//无法确定顺序
			printf("Sorted sequence cannot be determined.\n");
		} 
	}
	return 0;
}

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.

标签:Sorting,POJ1094,sequence,int,relations,次序,sorted,include,Out
来源: https://blog.csdn.net/weixin_43735161/article/details/100511302

本站声明: 1. iCode9 技术分享网(下文简称本站)提供的所有内容,仅供技术学习、探讨和分享;
2. 关于本站的所有留言、评论、转载及引用,纯属内容发起人的个人观点,与本站观点和立场无关;
3. 关于本站的所有言论和文字,纯属内容发起人的个人观点,与本站观点和立场无关;
4. 本站文章均是网友提供,不完全保证技术分享内容的完整性、准确性、时效性、风险性和版权归属;如您发现该文章侵犯了您的权益,可联系我们第一时间进行删除;
5. 本站为非盈利性的个人网站,所有内容不会用来进行牟利,也不会利用任何形式的广告来间接获益,纯粹是为了广大技术爱好者提供技术内容和技术思想的分享性交流网站。

专注分享技术,共同学习,共同进步。侵权联系[81616952@qq.com]

Copyright (C)ICode9.com, All Rights Reserved.

ICode9版权所有