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…

THE MATRIX PROBLEM Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5437    Accepted Submission(s): 1372 Problem Description You have been given a matrix CN*M, each element E of CN*M is positive…

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+…

题意: 给一个n*m矩阵,每个格子上有一个数字a[i][j],给定L和U,问:是否有这样两个序列{a1...an}和{b1...bn},满足 L<=a[i][j]*ai/bj<=U .若存在输出yes,否则no. 思路: 能够得到的是一堆不等式,那么可以用最短路来解决差分约束系统.但是a[i][j]*ai/bj<=U是除的,得提前变成减的才行.可以用对数log来解决,先不管a[i][j],logai-logbj<=U不就行了?可以得到: (1)logai - logbj<=U…

差分约束系统. 根据题意,可以写出不等式 L <= (Xij * Ai) / Bj <= U 即 Ai/Bj<=U/Xij和Ai/Bj>=L/Xij 由于差分约束系统是减法..除法变减法可以用对数来解决. 两个式子两边取对数,可以写成log(Ai)-log(Bj)<=log(U/Xij)和log(Ai)-log(Bj)>=log(L/Xij) log(Ai)和log(Bj)看作两个节点.编号分别为i和n+j,建立有向图,判断有没有负环存在.  if(summ[hh]&g…

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…

这道题有深搜和广搜.深搜还有要求,靠左或靠右.下面以靠左为例,可以把简单分为上北,下南,左西,右东四个方向.向东就是横坐标i不变,纵坐标j加1(i与j其实就是下标).其他方向也可以这样确定.通过上一步方向可以确定下一步应该从哪个方向开始搜.比如说,是向北走的,就必须先搜西,西不可以走,再搜北,如果北还不可以走,再搜东,最后才是南.其他方向的情况也可以这样推出来.最后走到E点完成了.广搜就是最基础的广搜.这道题做了将近10个小时.中途曾几次准备放弃,但最后还是坚持做完了. #include<ios…