POJ 1094 Sorting It All Out
2013-04-07 21:21
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题意:给你N个字母( 输入时保证是字母表的前n个 ), M个关系( 关系全部为 X<Y 的格式
输出的答案有三种:
1.根据所给的关系可以确定唯一的拓扑排序( 还要记录在给出的前几组就可以确定,不一定用到所有的关系 )
2.所给的关系有相互矛盾的(同样要记录出根据前几组就可以得出这个结论)
3.用上所有的关系仍然不能确定出拓扑排序
首先,这道题用拓扑排序毋庸置疑,但是在排序的时候要做一些修改,题目的难点也就是在这里
具体详见代码:
Sorting It All Out
Description
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
Sample Output
输出的答案有三种:
1.根据所给的关系可以确定唯一的拓扑排序( 还要记录在给出的前几组就可以确定,不一定用到所有的关系 )
2.所给的关系有相互矛盾的(同样要记录出根据前几组就可以得出这个结论)
3.用上所有的关系仍然不能确定出拓扑排序
首先,这道题用拓扑排序毋庸置疑,但是在排序的时候要做一些修改,题目的难点也就是在这里
具体详见代码:
Sorting It All Out
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 23043 | Accepted: 7965 |
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.
#include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> using namespace std; #define MAXN 27 int in[MAXN], temp[MAXN], topo[MAXN]; //入度,topo排序结果 bool visit[MAXN][MAXN]; //标记i, j存在i < j关系 int n, m;//字母数,边数 int toposort() //拓扑排序,三个返回值:环、继续读边(未完成)、排序完成 { int u, v, count; int num; //入度为0的字母 bool flag; count = 1; //已排序字母个数 flag = true; memset(topo, 0, sizeof(topo)); for(int i = 1; i <= n; ++i) temp[i] = in[i]; for(int i = 1; i <= n; ++i) { num = 0; //这个不能写在外面,删边后度变化。。。。错了N久啊。。。 for(int j = 1; j <= n; ++j) //统计入度为0的字母个数 if(temp[j] == 0) { u = j; num++; } if(num == 0) //为环 return 0; if(num > 1) //多条分支,继续读边 flag = false; temp[u] = -1; topo[count++] = u; //加入拓扑序列 for(int j = 1; j <= n; ++j) //删边操作 if(visit[u][j] == true) temp[j]--; } if(flag == false) return -1; return 1; //经过n次排序到达这里,则排序完成 } int main() { char str[5]; int start, end; bool flag; int res; //toposort返回的结果 while(scanf("%d%d", &n, &m) != EOF) { if(n == 0 && m == 0) break; memset(in, 0, sizeof(in)); memset(visit, false, sizeof(visit)); flag = true; for(int i = 1; i <= m; ++i) { scanf("%s", str); if(flag == false) //出现环或者矛盾 continue; start = str[0] - 'A' + 1; end = str[2] - 'A' + 1; in[end]++; visit[start][end] = true; res = toposort(); //加一次边拓扑排序一次 if(res == 0) { printf("Inconsistency found after %d relations.\n", i); flag = false; } else if(res == 1) { printf("Sorted sequence determined after %d relations: ", i); for(int j = 1; j <= n; ++j) printf("%c", topo[j] + 'A' - 1); printf(".\n"); flag = false; } } if(flag == true) printf("Sorted sequence cannot be determined.\n"); } return 0; }
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