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…

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…

King Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 1645    Accepted Submission(s): 764 Problem Description Once, in one kingdom, there was a queen and that queen was expecting a baby. The quee…

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…

Problem - 1384 好歹用了一天,也算是看懂了差分约束的原理,做出第一条查分约束了. 题意是告诉你一些区间中最少有多少元素,最少需要多少个元素才能满足所有要求. 构图的方法是,(a)->(b+1)=c.还有就是所有的相邻的点都要连上(i+1)->(i)=0,(i)->(i+1)=-1.因为我对点离散了,所以就变成(rx[i])->(rx[i+1])=rx[i]-rx[i+1]. 代码如下: #include <cstdio> #include <cstr…

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+…

类型:给出一些形如a−b<=k的不等式(或a−b>=k或a−b<k或a−b>k等),问是否有解[是否有负环]或求差的极值[最短/长路径].例子:b−a<=k1,c−b<=k2,c−a<=k3.将a,b,c转换为节点:k1,k2,k3转换为边权:减数指向被减数,形成一个有向图: 由题可得(b−a) + (c−b) <= k1+k2,c−a<=k1+k2.比较k1+k2与k3,其中较小者就是c−a的最大值.由此我们可以得知求差的最大值,即上限被约束,此时我…

// 题目描述:一个项目被分成几个部分,每部分必须在连续的天数完成.也就是说,如果某部分需要3天才能完成,则必须花费连续的3天来完成它.对项目的这些部分工作中,有4种类型的约束:FAS, FAF, SAF和SAS.两部分工作之间存在一个FAS约束的含义是:第一部分工作必须在第二部分工作开始之后完成: Xa+Ta>=XbFAF约束的含义是:第一部分工作必须在第二部分工作完成之后完成: Xa+Ta>=Xb+TbSAF的含义是:第一部分工作必须在第二部分工作完成之后开始: Xa>=Xb+TbS…