poj 3169&hdu3592(差分约束)】的更多相关文章

Layout Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9687   Accepted: 4647 Description Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a…
            布局 题目大意:有N头牛,编号1-N,按编号排成一排准备吃东西,有些牛的关系比较好,所以希望他们不超过一定的距离,也有一些牛的关系很不好,所以希望彼此之间要满足某个关系,牛可以挤在同一个位置上,现在给出N个牛的信息,问你能否实现一种排列方案,使得d[1]到d[N]最大?如果不存在输出-1,无限大输出-2 这一题看上去挺难的,但是如果你知道差分约束原理,这一题似乎还是挺简单的. 差分约束的原理是:存在任意线性方程,满足d[A]+c>=d[B],就可以表示为图的最短路形式,方…
题意:给定一些母牛,要求一个排列,有的母牛距离不能超过w,有的距离不能小于w,问你第一个和第n个最远距离是多少. 析:以前只是听说过个算法,从来没用过,差分约束. 对于第 i 个母牛和第 i+1 个,D[i] - D[i+1] <= 0,  D[j] -D[i ]<= k, D[i] - D[j] <= - k,那么这个题就可以用差分约束来求这个不等式组了. 1.对于差分不等式,a - b <= c ,建一条 b 到 a 的权值为 c 的边,求的是最短路,得到的是最大值(本题求的就…
描述 Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standing in the same order as they are numbered, and…
题目链接:http://poj.org/problem? id=3169 Description Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a straight line waiting for feed. The cows are standin…
题意: 有N头牛, 有以下关系: (1)A牛与B牛相距不能大于k (2)A牛与B牛相距不能小于k (3)第i+1头牛必须在第i头牛前面 给出若干对关系(1),(2) 求出第N头牛与第一头牛的最长可能距离, 若无解输出-1, 若无限长输出-2 分析: 3个关系对应的 <= 式子是: dis[b] - dis[a] <= d(1) dis[a] - dis[b] <= -d(2) dis[i] - dis[i+1] <= -1(2) 目标式:dis[N] - dis[1] <=…
Layout Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6415   Accepted: 3098 Description Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N standing along a…
题目:http://poj.org/problem?id=1201 题意:给定n组数据,每组有ai,bi,ci,要求在区间[ai,bi]内至少找ci个数, 并使得找的数字组成的数组Z的长度最小. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cstdlib> using namespace std; <<;…
题目:http://poj.org/problem?id=1201 差分约束裸套路:前缀和 本题可以不把源点向每个点连一条0的边,可以直接把0点作为源点.这样会快许多! 可能是因为 i-1 向 i 都连着一条0的边. 别忘了约束条件不仅有s[ i ] - s[ i-1 ] >= 0,还有s[ i ] - s[ i - 1] <= 1! 别忘了s的范围不是n而是mx! #include<iostream> #include<cstdio> #include<cstr…
POJ——3169Layout Layout Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14702   Accepted: 7071 Description Like everyone else, cows like to stand close to their friends when queuing for feed. FJ has N (2 <= N <= 1,000) cows numbered 1..N…