点击打开链接 Redraw Beautiful Drawings Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 1660    Accepted Submission(s): 357 Problem Description Alice and Bob are playing together. Alice is crazy about…

Problem Description Alice and Bob are playing together. Alice is crazy about art and she has visited many museums around the world. She has a good memory and she can remember all drawings she has seen. Today Alice designs a game using these drawings…

题目 Source http://acm.hdu.edu.cn/showproblem.php?pid=4888 Description Alice and Bob are playing together. Alice is crazy about art and she has visited many museums around the world. She has a good memory and she can remember all drawings she has seen.…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4888 题意:给一个矩阵没行的和和每列的和,问能否还原矩阵,如果可以还原解是否唯一,若唯一输出该矩阵. 思路:设一个源点和汇点,每行的和和源点加边,权值为该行的和,每列的和和汇点加点,权值为该列的和. 每行和每列加边, 权值为k,跑最大流,如果满流(从源点流出的等于流入汇点的)则证明可以还原该矩阵.   判断是否有多节,就dfs搜,看残余网络中是否有环,无环则解唯一. AC代码: #include<i…

传送门:pid=4888">[HDU]4888 Redraw Beautiful Drawings 题目分析: 比赛的时候看出是个网络流,可是没有敲出来.各种反面样例推倒自己(究其原因是不愿意写暴力推断的).. 首先是简单的行列建边.源点向行建边.容量为该行元素和,汇点和列建边.容量为该列元素和.全部的行向全部的列建边,容量为K. 跑一次最大流.满流则有解,否则无解. 接下来是推断解是否唯一. 这个题解压根没看懂.还是暴力大法好. 最简单的思想就是枚举在一个矩形的四个端点.设A.D为主对角…

Redraw Beautiful Drawings Time Limit: 3000/1500 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 2909 Accepted Submission(s): 942 Problem Description Alice and Bob are playing together. Alice is crazy about art and she h…

hdu4888 Redraw Beautiful Drawings Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 2007    Accepted Submission(s): 447 Problem Description Alice and Bob are playing together. Alice is crazy abou…

Redraw Beautiful Drawings Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Problem Description Alice and Bob are playing together. Alice is crazy about art and she has visited many museums around the world. She has…

pid=4888">http://acm.hdu.edu.cn/showproblem.php?pid=4888 加入一个源点与汇点,建图例如以下: 1. 源点 -> 每一行相应的点,流量限制为该行的和 2. 每一行相应的点 -> 每一列相应的点,流量限制为 K 3. 每一列相应的点 -> 汇点,流量限制为该列的和 求一遍最大流,若最大流与矩阵之和相等,说明有解,否则无解.推断唯一解,是推断残量网络中是否存在一个长度大于2的环.若存在说明有多解,否则有唯一解,解就是每条边…

好难好难,将行列当成X和Y,源汇点连接各自的X,Y集,容量为行列的和,相当于从源点流向每一行,然后分配流量给每一列,最后流入汇点,这样执意要推断最后是否满流,就知道有没有解,而解就是每一行流向每一列多少流量. 关键在于怎么推断多解的情况.我想不到啊T_T 题讲解,找到一个长度大于2的环. 想了一想,也就是找到还有剩余流量的环,假设找到了,我就能够把当中一条边的流量转移,由于是一个环,所以它又会达到平衡,不会破坏最大流,可是这样转移后,解就多了一种,所以仅仅要推断是否有一个长度大于2的环就够了.…