http://poj.org/problem?id=1505 Copying Books Time Limit: 3000MS   Memory Limit: 10000K Total Submissions: 7053   Accepted: 2200 Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be r…

此题主要采用DP思路,难点是求解"/",需要考虑划分数量不够的情况,先采用DP求解最优解,然后采用贪心求解slash.防止中间结果出错,使用了unsigned int. #include <iostream> using namespace std; #define MAXNUM 501 unsigned int pages[MAXNUM][MAXNUM]; int main() { int case_n, m, k; int i, j, p, q; unsigned mi…

题目链接: 传送门 Copying Books Time Limit: 3000MS     Memory Limit: 32768 KB Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had b…

Copying  Books 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=85904#problem/B 题目: Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so calledscri…

题目链接: 题目 Copying Books Time limit: 3.000 seconds 问题描述 Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after s…

题目连接:714 - Copying Books 题目大意:将一个个数为n的序列分割成m份,要求这m份中的每份中值(该份中的元素和)最大值最小, 输出切割方式,有多种情况输出使得越前面越小的情况. 解题思路:二分法求解f(x), f(x) < 0 说明不能满足, f(x) >= 0说明可以满足,f(x) 就是当前最大值为x的情况最少需要划分多少份-要求份数(如果f(x ) >= 0 说明符合要求而且还过于满足,即x还可以更小). 注意用long long . #include <s…

  Copying Books  Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after several months he finished its copy. O…

Copying Books Time Limit: 3000MS Memory Limit: 10000K Total Submissions: 7109 Accepted: 2221 Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scri…

Copying Books 给出一个长度为m的序列\(\{a_i\}\),将其划分成k个区间,求区间和的最大值的最小值对应的方案,多种方案,则按从左到右的区间长度尽可能小(也就是从左到右区间长度构成的序列的字典序最小),\(m,k\leq 500\). 解 显然最大值的最小值想到二分,其实dp也可以,因为区间划分问题有可递推性,而且它能求出答案. 二分一个东西就等于换时间复杂度增加个\(log(n)\)增加了一个已知条件,虽然前提是单调性,但不妨以后缺少条件,先考虑二分再找单调性. 我们二分\(…

Before the invention of book-printing, it was very hard to make a copy of a book. All the contents hadto be re-written by hand by so called scribers. The scriber had been given a book and after severalmonths he finished its copy. One of the most famo…

Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Description   Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so calledscribers. T…

B - 二分 Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu   Description   Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so calledscribers. The sc…

Description Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after several months he finished its copy. One of…

Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after several months he finished its copy. One of the most fa…

首先通过二分来确定这种最大值最小的问题. 假设每个区间的和的最大值为x,那么只要判断的时候只要贪心即可. 也就是如果和不超过x就一直往区间里放数,否则就开辟一个新的区间,这样来判断是否k个区间容得下这些数. 还有就是输出也挺麻烦的,借鉴了一下lrj的代码,感觉也是十分巧妙. #include <bits/stdc++.h> using namespace std; typedef long long LL; + ; LL a[maxn]; bool last[maxn]; int n, k;…

题目描写叙述开头一大堆屁话,我还细致看了半天..事实上就最后2句管用.意思就是给出n本书然后要分成k份,每份总页数的最大值要最小.问你分配方案,假设最小值同样情况下有多种分配方案,输出前面份数小的,就像字典序输出从小到大一样的意思. 这里用到贪心的方法,定义f(x)为真的条件是满足x为最大值使n本书分成k份,那么就是求x的最小值.怎样确定这个x就是用的二分法,x一定大于0小于全部值的合,不断的二分再推断是否成立,成立就取左半边,不成立说明太小了就取右半边,写的时候还是没有把二分法理解透彻,我还怕…

题意: 要抄N本书,编号为1,2,3...N, 每本书有1<=x<=10000000页, 把这些书分配给K个抄写员,要求分配给某个抄写员的那些书的编号必须是连续的.每个抄写员的速度是相同的,求所有书抄完所用的最少时间的分配方案. 分析: 这个题以前做过.就是先二分出来,最大的区间最小值.然后一重循环查找输出/就好 代码: #include <iostream>#include <cstring>#include <cstdio>#include <al…

传送门:Zoj2002 题目大意:从左到右把一排数字k分,得到最小化最大份,如果有多组解,左边的尽量小. 思路:贪心+二分(参考青蛙过河). 方向:从右向左. 注意:有可能最小化时不够k分.如                                     3 3                            1 2 3   k分得到最小化的最大值是3,分组却只能分两个组.    错误结果是 1 2 / 3.正确结果是1 / 2 / 3   因此要从左到右补齐'/' #inclu…

求使最大值最小,可以想到二分答案. 然后再根据题目意思乱搞一下,按要求输出斜杠(这道题觉得就这一个地方难). Code /** * UVa * Problem#12627 * Accepted * Time:0ms */ #include<iostream> #include<cstdio> #include<cctype> #include<ctime> #include<cstring> #include<cstdlib> #in…

好久不更新主要是怠惰了....还要加强训练. 题意分析与思路 注意到这样一句话: our goal is to minimize the maximum number of pages assigned to a single scriber. Your task is to find the optimal assignment. 这种最大化最小.最小化最大,显然是二分. 如何二分呢,枚举分成k份中各份的最大值,判断在$max_t$的情况下能否分成$\le k$份,能的话那么我们的$max_t…

题目大意: 要抄N本书,编号为1,2,3...N, 每本书有1<=x<=10000000页, 把这些书分配给K个抄写员,要求分配给某个抄写员的那些书的编号必须是连续的.每个抄写员的速度是相同的,求所有书抄完所用的最少时间的分配方案. 题目中的要求是去求划分的子序列的最大值尽量小,最大值最小化,如果从划分的角度看,无法获得好的思路,我们可以从值得角度考虑,所要求的最小的最大值必定是从[amax,sum(总和)]中取得的,那么我们可以二分法的方式猜测一个数字,看它是否满足要求,如果满足要求,我们可…

题意:把一个包含m个正整数的序列划分成k个非空的连续子序列.使得所有连续子序列的序列和Si的最大值尽量小. 二分,每次判断一下当前的值是否满足条件,然后修改区间.注意初始区间的范围,L应该为所有正整数中的最大值,否则应该判断时注意.输出解的时候要使字典序最小,所以从后面贪心. #include<bits/stdc++.h> using namespace std; typedef long long ll; ; ll p[maxm]; bool vis[maxm]; int k,m; inli…

[链接] 我是链接,点我呀:) [题意] 在这里输入题意 [题解] 二分最后的最大值的最小值. 得到ans 然后从后往前尽量划分. 如果发现不够分成k个. 那么就从第一个开始接着分restk个(每隔1个分1块 中间遇到之前分了的就直接跳过 [代码] /* 1.Shoud it use long long ? 2.Have you ever test several sample(at least therr) yourself? 3.Can you promise that the soluti…

题意:把一个包含m个正整数的序列划分成k个(1<=k<=m<=500)非空的连续子序列,使得每个正整数恰好属于一个序列(所有的序列不重叠,且每个正整数都要有所属序列).设第i个序列的各数之和为S(i),你的任务是让所有的S(i)的最大值尽量小.如果有多解,S(1)应尽量小,如果仍有多解,S(2)应尽量小,依此类推. 分析: 1.二分最小值x. 2.判断当前x是否满足条件时,从右往左尽量划分,若cnt<k,则从0开始依次标为分界点,这样可满足S(1),S(2),……,尽量小. #pr…

题意:将1个含N个正整数的序列划分成K个连续的子序列,使每段的和的最大值尽量小,问字典序最小的划分方案. 解法:由于是连续的数的"最大值最小",便可想到二分每段的最大值,若这时可分成<=K段,则这个最大值成立,再继续二分. 输出方案需要用到贪心策略, 先从后往前贪心求得最小划分的段数M,若M不足K,则直接使K-M个数单独划分为一段,保证字典序最小. 1 #include<cstdio> 2 #include<cstdlib> 3 #include<c…

此文转载别人,希望自己能够做完这些题目! 1.POJ动态规划题目列表 容易:1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276,1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈),1742, 1887, 1926(马尔科夫矩阵,求平衡), 1936, 1952, 1953, 1958, 1959, 1962, 1975,…

[1]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈), 1742, 1887,1926(马尔科夫矩阵,求平衡), 1936, 1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029,…

]POJ 动态规划题目列表 容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈), 1742, 1887, 1926(马尔科夫矩阵,求平 衡), 1936,1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029,2…

此文转载别人,希望自己可以做完这些题目. 1.POJ动态规划题目列表 easy:1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1208, 1276,1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740(博弈),1742, 1887, 1926(马尔科夫矩阵,求平衡), 1936, 1952, 1953, 1958, 1959, 1962, 1975…

列表一:经典题目题号:容易: 1018, 1050, 1083, 1088, 1125, 1143, 1157, 1163, 1178, 1179, 1189, 1191,1208, 1276, 1322, 1414, 1456, 1458, 1609, 1644, 1664, 1690, 1699, 1740, 1742, 1887, 1926, 1936, 1952, 1953, 1958, 1959, 1962, 1975, 1989, 2018, 2029, 2039, 2063, 20…