题目请戳这里 题目大意:给一个n*m的矩阵,求是否存在这样两个序列:a1,a2...an,b1,b2,...,bm,使得矩阵的第i行乘以ai,第j列除以bj后,矩阵的每一个数都在L和U之间. 题目分析:比较裸的差分约束.考虑那2个序列,可以抽象出m+n个点.乘除法可以通过取对数转换为加减法.然后就可以得到约束关系: 对于矩阵元素cij,有log(L) <= log(cij) + ai - bj <= log(U),整理可得: ai - bj <= log(U) - log(cij),n+…

You have been given a matrix C N*M, each element E of C N*M is positive and no more than 1000, The problem is that if there exist N numbers a1, a2, … an and M numbers b1, b2, …, bm, which satisfies that each elements in row-i multiplied with ai and e…

You have been given a matrix C N*M, each element E of C N*M is positive and no more than 1000, The problem is that if there exist N numbers a1, a2, - an and M numbers b1, b2, -, bm, which satisfies that each elements in row-i multiplied with ai and e…

题意:给定一个最大400*400的矩阵,每次操作可以将某一行或某一列乘上一个数,问能否通过这样的操作使得矩阵内的每个数都在[L,R]的区间内. 析:再把题意说明白一点就是是否存在ai,bj,使得l<=cij*(ai/bj)<=u (1<=i<=n,1<=j<=m)成立. 首先把cij先除到两边去,就变成了l'<=ai/bj<=u',由于差分约束要是的减,怎么变成减法呢?取对数呗,两边取对数得到log(l')<=log(ai)-log(bj)<=l…

THE MATRIX PROBLEM Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 8693    Accepted Submission(s): 2246 Problem Description You have been given a matrix CN*M, each element E of CN*M is positive…

Schedule Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1085    Accepted Submission(s): 448Special Judge Problem Description A project can be divided into several parts. Each part shoul…

差分约数: 求满足不等式条件的尽量小的值---->求最长路---->a-b>=c----> b->a (c) Schedule Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 1503    Accepted Submission(s): 647 Special Judge Problem Descr…

转自:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548 A strange lift基础最短路(或bfs)★2544 最短路 基础最短路★3790 最短路径问题基础最短路★2066 一个人的旅行基础最短路(多源多汇,可以建立超级源点和终点)★2112 HDU Today基础最短路★1874 畅通工程续基础最短路★1217 Arbitrage 货币交换 Floyd (或者 Bellman-Ford 判环)★124…

出处:http://blog.csdn.net/shahdza/article/details/7779273 最短路 [HDU] 1548    A strange lift基础最短路(或bfs)★ 2544    最短路  基础最短路★ 3790    最短路径问题基础最短路★ 2066    一个人的旅行基础最短路(多源多汇,可以建立超级源点和终点)★ 2112    HDU Today基础最短路★ 1874    畅通工程续基础最短路★ 1217    Arbitrage   货币交换…

怎么搞?        1. 如果要求最大值      想办法把每个不等式变为标准x-y<=k的形式,然后建立一条从y到x权值为k的边,变得时候注意x-y<k =>x-y<=k-1     将这些约束条件转化为差分约束,不妨设T[x] = S[1]+S[2]+....S[x],那么上面式子就可以转化为:         1. T[si+ni] - T[si-1] > ki          2. T[si+ni] - T[si-1] < ki          又差分约…