码迷,mamicode.com
首页 > 其他好文 > 详细

POJ 1094 Sorting It All Out

时间:2014-09-12 01:06:42      阅读:178      评论:0      收藏:0      [点我收藏+]

标签:style   blog   http   color   io   os   java   ar   for   

Sorting It All Out

Time Limit: 1000ms
Memory Limit: 10000KB
This problem will be judged on PKU. Original ID: 1094
64-bit integer IO format: %lld      Java class name: Main
 
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.

Source

 
解题:这题说的是边读入一些关系。如果当前出现矛盾的(有环),就记录出现矛盾的是第几组关系。如果当前组已经能够确定n个字母的关系,也要记录当前是第几组。如果当前关系不唯一。那么继续了。
 
判关系是唯一就是时刻判断队列中的元素是不是大于1.如果大于1,说明同时有几个点入度为0.排序就不确定了。
 
bubuko.com,布布扣
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 #include <algorithm>
 6 #include <climits>
 7 #include <vector>
 8 #include <queue>
 9 #include <cstdlib>
10 #include <string>
11 #include <set>
12 #include <stack>
13 #define LL long long
14 #define pii pair<int,int>
15 #define INF 0x3f3f3f3f
16 using namespace std;
17 const int maxn = 105;
18 int n,m,cnt,in[maxn],tmp[maxn],result[maxn];
19 vector<int>g[maxn];
20 queue<int>q;
21 void init() {
22     for(int i = 0; i < maxn; i++) {
23         in[i] = 0;
24         g[i].clear();
25     }
26 }
27 bool Find(int x,int y) {
28     for(int i = 0; i < g[x].size(); i++)
29         if(g[x][i] == y) return false;
30     return true;
31 }
32 int topSort() {
33     while(!q.empty()) q.pop();
34     for(int i = 0; i < n; i++)
35         if(!in[i]) q.push(i);
36     cnt = 0;
37     bool uncertain = false;
38     while(!q.empty()) {
39         int u = q.front();
40         if(q.size() > 1) uncertain = true;
41         q.pop();
42         result[cnt++] = u;
43         for(int i = 0; i < g[u].size(); i++) {
44             if(--in[g[u][i]] == 0) q.push(g[u][i]);
45         }
46     }
47     if(cnt < n) return 1;
48     if(uncertain) return 2;
49     return 3;
50 }
51 int main() {
52     bool ok;
53     char u,v,op;
54     int endPoint,flag;
55     while(scanf("%d %d",&n,&m),n||m) {
56         init();
57         ok = false;
58         getchar();
59         flag = 2;
60         for(int i = 1; i <= m; i++) {
61             scanf("%c%c%c%*c",&u,&op,&v);
62             if(ok) continue;
63             if(op == < && Find(v-A,u-A)) {
64                 g[v-A].push_back(u-A);
65                 in[u-A]++;
66             } else if(op == > && Find(u-A,v-A)) {
67                 g[u-A].push_back(v-A);
68                 in[v-A]++;
69             }
70             memcpy(tmp,in,sizeof(in));
71             flag = topSort();
72             if(flag != 2) {
73                 endPoint = i;
74                 ok = true;
75             }
76             memcpy(in,tmp,sizeof(in));
77         }
78         if(flag == 3) {
79             printf("Sorted sequence determined after %d relations: ", endPoint);
80             for(int i = cnt-1; i >= 0; i--)
81                 printf("%c",result[i]+A);
82             puts(".");
83         } else if(flag == 1) printf("Inconsistency found after %d relations.\n",endPoint);
84         else puts("Sorted sequence cannot be determined.");
85     }
86     return 0;
87 }
View Code

 

POJ 1094 Sorting It All Out

标签:style   blog   http   color   io   os   java   ar   for   

原文地址:http://www.cnblogs.com/crackpotisback/p/3967479.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!