• [题解] poj 1094 Sorting It All Out (拓扑排序)


    - 传送门 -

     http://poj.org/problem?id=1094

    #Sorting It All Out

    | Time Limit: 1000MS |   | Memory Limit: 10000K |
    | Total Submissions: 35759 |   | Accepted: 12590 |

    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

    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

    East Central North America 2001

    - 题意 -

     有 n 个大写字母 (A 到 A + n - 1), m 组大小关系.
     设到第 k 组的时候能确定唯一的大小关系或确定发生了冲突, 求这个 k 值, 或是输完 m 组也没有冲突, 也不能确定唯一顺序.
     

    - 思路 -

     好模板哦!
     注意一下三种情况的判断顺序就好.
     
     细节见代码.
     

    - 代码 -

    #include<cstdio>
    #include<cstring>
    #include<queue>
    using namespace std;
    
    const int N = 30;
    
    int ANS[N], TMP[N], IN[N], MAP[N][N];
    int n, m, ans, sz, f;
    char S[10];
    queue<int> q;
    
    int topo() {
    	sz = 0, f = 0;
    	while (!q.empty())
    		q.pop();
    	for (int i = 1; i <= n; ++i)
    		if (TMP[i] == 0) {
    			q.push(i);
    			TMP[i]--;
    		}
    	while (!q.empty()) {
    		if (q.size() > 1) f = -1;
    		int i = q.front();
    		q.pop();
    		ANS[++sz] = i;
    		for (int j = 1; j <= n; ++j) {
    			if (MAP[i][j]) TMP[j]--;
    			if (TMP[j] == 0) {
    				q.push(j);
    				TMP[j]--;
    			}
    		}
    	}
    	for (int i = 1; i <= n; ++i)
    		if (TMP[i] != -1)
    			return -1;
    	if (f == -1) return 0;
    	return 1; //注意顺序, 先判环的存在, 再判无法确定顺序的情况.
    }
    
    int main() {
    	while(~scanf("%d%d", &n, &m) && n && m) {
    		memset(IN, 0, sizeof (IN));
    		memset(MAP, 0, sizeof (MAP));
    		ans = 0;
    		for (int i = 1; i <= m; ++i) {
    			scanf("%s", S);
    			if (ans != 0) continue;
    			int a = S[0] - 'A' + 1, b = S[2] - 'A' + 1;
    			if (!MAP[b][a]) IN[a] ++;
    				MAP[b][a] = 1;
    			for (int j = 1; j <= n; ++j)
    				TMP[j] = IN[j];
    			ans = topo();
    			if (ans == 1) {
    				printf("Sorted sequence determined after %d relations: ", i);
    				for (int j = n; j >= 1; --j)
    					printf("%c", ANS[j] + 'A' - 1);
    				printf(".
    ");
    			}
    			if (ans == -1)
    				printf("Inconsistency found after %d relations.
    ", i);
    		}
    		if (ans == 0)
    			printf("Sorted sequence cannot be determined.
    ");
    	}
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Anding-16/p/7402787.html
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