Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 605    Accepted Submission(s): 127 Problem Description It's really a simple problem.  Given a "01" matrix with size by n*n (the matrix size…

01 Matrix 题解 原创文章,拒绝转载 题目来源:https://leetcode.com/problems/01-matrix/description/ Description Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell. The distance between two adjacent cells is 1. Example 1: Input: 0 0 0 0…

01 Matrix Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 564    Accepted Submission(s): 121 Problem Description It's really a simple problem.  Given a "01" matrix with size by n*n (the ma…

HDU 4920 Matrix multiplication 题目链接 题意:给定两个矩阵,求这两个矩阵相乘mod 3 思路:没什么好的想法,就把0的位置不考虑.结果就过了.然后看了官方题解,上面是用了bitset这个东西,能够用来存大的二进制数,那么对于行列相乘.事实上就几种情况,遇到0都是0了,1 1得1,2 1,1 2得2,2 2得1.所以仅仅要存下行列1和2存不存在分别表示的二进制数.然后取且bitcount一下的个数,就能够计算出对应的数值了 代码: 暴力: #include <cst…

HDU 2686 Matrix 题目链接 3376 Matrix Again 题目链接 题意:这两题是一样的,仅仅是数据范围不一样,都是一个矩阵,从左上角走到右下角在从右下角走到左上角能得到最大价值 思路:拆点.建图,然后跑费用流就可以,只是HDU3376这题,极限情况是300W条边,然后卡时间过了2333 代码: #include <cstdio> #include <cstring> #include <vector> #include <queue>…

给予一个矩阵,矩阵有1有0,计算每一个1到0需要走几步,只能走上下左右. 解法一: 利用dp,从左上角遍历一遍,再从右下角遍历一遍,dp存储当前位置到0的最短距离. 十分粗心的搞错了col和row,改了半天………… Runtime: 132 ms, faster than 98.88% of C++ online submissions for 01 Matrix. class Solution { public: vector<vector<int>> updateMatrix(…

542. 01 Matrix https://www.cnblogs.com/grandyang/p/6602288.html 将所有的1置为INT_MAX,然后用所有的0去更新原本位置为1的值. 最短距离肯定使用bfs. 每次更新了值的地方还要再加入队列中 . class Solution { public: vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) { ].size…

[LeetCode]01 Matrix 解题报告 标签(空格分隔): LeetCode 题目地址:https://leetcode.com/problems/01-matrix/#/description 题目描述: Given a matrix consists of 0 and 1, find the distance of the nearest 0 for each cell. The distance between two adjacent cells is 1. Example:…

题目地址http://acm.hdu.edu.cn/showproblem.php?pid=2602 #include <stdio.h> #include <string.h> int main() { ],f[],va[]; int i,j,k,n,m; scanf("%d",&k); while(k--) { memset(f,,sizeof(f)); scanf("%d%d",&n,&m); ;i<n;i…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2686 Yifenfei very like play a number game in the n*n Matrix. A positive integer number is put in each area of the Matrix.Every time yifenfei should to do is that choose a detour which frome the top left…