标签:prerequisites course int disto Schedule IV Course queries true
There are a total of n
courses you have to take, labeled from 0
to n-1
.
Some courses may have direct prerequisites, for example, to take course 0 you have first to take course 1, which is expressed as a pair: [1,0]
Given the total number of courses n
, a list of direct prerequisite
pairs and a list of queries
pairs.
You should answer for each queries[i]
whether the course queries[i][0]
is a prerequisite of the course queries[i][1]
or not.
Return a list of boolean, the answers to the given queries
.
Please note that if course a is a prerequisite of course b and course b is a prerequisite of course c, then, course a is a prerequisite of course c.
Example 1:
Input: n = 2, prerequisites = [[1,0]], queries = [[0,1],[1,0]] Output: [false,true] Explanation: course 0 is not a prerequisite of course 1 but the opposite is true.
Example 2:
Input: n = 2, prerequisites = [], queries = [[1,0],[0,1]] Output: [false,false] Explanation: There are no prerequisites and each course is independent.
Example 3:
Input: n = 3, prerequisites = [[1,2],[1,0],[2,0]], queries = [[1,0],[1,2]] Output: [true,true]
Example 4:
Input: n = 3, prerequisites = [[1,0],[2,0]], queries = [[0,1],[2,0]] Output: [false,true]
Example 5:
Input: n = 5, prerequisites = [[0,1],[1,2],[2,3],[3,4]], queries = [[0,4],[4,0],[1,3],[3,0]] Output: [true,false,true,false]
Constraints:
2 <= n <= 100
0 <= prerequisite.length <= (n * (n - 1) / 2)
0 <= prerequisite[i][0], prerequisite[i][1] < n
prerequisite[i][0] != prerequisite[i][1]
- The prerequisites graph has no cycles.
- The prerequisites graph has no repeated edges.
1 <= queries.length <= 10^4
queries[i][0] != queries[i][1]
class Solution { public List<Boolean> checkIfPrerequisite(int n, int[][] prerequisites, int[][] queries) { List<Boolean> res = new ArrayList(); int[][] disto = new int[n][n]; for(int i = 0; i < n; i++){ Arrays.fill(disto[i], 1000); disto[i][i] = 0; } for(int[] arr: prerequisites){ disto[arr[0]][arr[1]] = 1; } for (int k = 0; k < n; k++) { for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (disto[i][j] > disto[i][k] + disto[k][j]) disto[i][j] = disto[i][k] + disto[k][j]; } } } for(int i = 0; i < queries.length; i++){ if(disto[queries[i][0]][queries[i][1]] == 1000) res.add(false); else res.add(true); } return res; } }
floyd-walshall算法:用二维数组存放i 到 j 经过 k所需要的最小cost。
这题就按正常先把数组设置好,然后就从query里查能不能从i到j即可(等于初始值就不可以)
标签:prerequisites,course,int,disto,Schedule,IV,Course,queries,true 来源: https://www.cnblogs.com/wentiliangkaihua/p/13029961.html
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