Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top. FJ has N cows (1 ≤ N ≤ 20) each with some height of Hi (1 ≤ Hi ≤ 1,000,000 - these ar…

http://poj.org/problem?id=3628 01背包 #include <cstdio> #include <iostream> #include <cstring> #include <algorithm> #define maxn 21 #define ll long long using namespace std; ll h[maxn]; int n; ll b; ll dp[]; ll max1(ll a,ll b) { retu…

传送门:http://poj.org/problem?id=3628 题目看了老半天,牛来叠罗汉- -|||和书架什么关系啊.. 大意是:一群牛来叠罗汉,求超过书架的最小高度. 0-1背包的问题,对于第i只牛可以放或者不放.然后最后求出大于书架高度的,减去书架高度即可. 也可以倒着来看.背包的容量为牛总的高度-书架的高度,求不超过这个容量的最大值,最后容量-这个值就是答案了.(推荐) 还可以DFS.. #include<cstdio> #include<cstring> #incl…

Bookshelf 2 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9488   Accepted: 4311 Description Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available…

Bookshelf 2 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7496   Accepted: 3451 Description Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available…

Bookshelf 2 Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11105   Accepted: 4928 Description Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only availabl…

题意:给出n头牛的身高,以及一个书架的高度,问怎样选取牛,使得它们的高的和超过书架的高度最小. 将背包容量转化为所有牛的身高之和,就可以用01背包来做=== #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #define maxn 2000005 using namespace std; ]; int main() { ,tmp=; scanf("%…

题目大意:FJ有n头奶牛,和一个高为h的架子,给出每头奶牛高度,求使奶牛叠加起来超过架子的最低高度是多少. 题目思路:求出奶牛叠加能达到的所有高度,并用dp[]保存,最后进行遍历,找出与h差最小的dp[]即所求答案. #include<cstdio> #include<stdio.h> #include<cstdlib> #include<cmath> #include<iostream> #include<algorithm> #i…

本题解法非常多,由于给出的数据特殊性故此能够使用DFS和BFS,也能够使用01背包DP思想来解. 由于一般大家都使用DFS,这里使用非常少人使用的BFS.缺点是比DFS更加耗内存,只是长处是速度比DFS快. 当然也比DFS难写点: int N, B; int Height[21]; inline int mMin(int a, int b) { return a > b? b : a; } inline int mMax(int a, int b) { return a < b? b : a;…

好久没看背包题目了!!!生疏了!!!! 这题是背包题!!!不过对于这题,解决方法还是搜索省时!!! 题意:第一行给你一个N和VV,接下来N行,每行一个数,求得是任选N个数组合求和,求组合的和大于VV而且减去VV的最小的差!!! 囧!!! ￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥￥ #include<stdio.h> #include<string.h> #include<string.h>…

http://poj.org/problem?id=3628 题意:给出一个高度H和n个牛的高度,要求把牛堆叠起来达到H,求出该高度和H的最小差. 思路:首先我们计算出牛的总高度sum,sum-H就相当于一个背包容量,如果我们往里装高度正好等于了sum-H,也就是说明我们堆叠的牛的高度正好等于了H. 这样一来很好理解,就是计算在一个背包容量为sum-H的背包中最多能装多少.题目本身还是不难的,直接套用模板就行了. #include<iostream> #include<algorithm…

思路: 1.01背包 先找到所有奶牛身高和与B的差. 然后做一次01背包即可 01背包的容积和价格就是奶牛们身高. 最后差值一减输出结果就大功告成啦! 2. 搜索 这思路很明了吧... 搜索的确可以过- 3. 模拟! 0到1< < n 来一遍.(状压呗) 01背包的: // by SiriusRen #include <cstdio> #include <algorithm> using namespace std; int N,B,sum=0,h[25],f[1000…

Description 有两棵APP树,编号为1,2.每一秒,这两棵APP树中的其中一棵会掉一个APP.每一秒,你可以选择在当前APP树下接APP,或者迅速移动到另外一棵APP树下接APP(移动时间可以忽略不计),但由于却乏锻炼,你最多移动W次.问在T秒内,你最多能收集多少个APP.假设你开始站在1号APP树下. Input 第1行:两个整数T(1 < = T< = 1000)和W(1 < = W< = 30)第2..T+1行:1或2,代表每分钟掉落APP的那棵树的编号 Outpu…

最近学业上堕落成渣了.得开始好好学习了. 还有呀,相家了,好久没回去啦~ 还有和那谁谁谁... 嗯,不能发表悲观言论.说好的. 如果这么点坎坷都过不去的话,那么这情感也太脆弱. ----------------------------------------------------好好学习的分割线---------------------------------------------------- poj:http://poj.org/problem?id=3627 大意: 给一些东西的高度,…

Description Farmer John recently bought another bookshelf for the cow library, but the shelf is getting filled up quite quickly, and now the only available space is at the top. FJ has N cows (1 ≤ N ≤ 20) each with some height of Hi (1 ≤ Hi ≤ 1,000,00…

Bookshelf Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7758   Accepted: 3906 Description Farmer John recently bought a bookshelf for cow library, but the shelf is getting filled up quite quickly, and now the only available space is at…

Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he…

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…

http://poj.org/problem?id=3628 就是比原题多了一个要求,输出>=m的最小值 kisang~独立做出来的都开心<(￣︶￣)> #include<cstdio> #include<cstring> #include<cmath> #include<iostream> #include<algorithm> using namespace std; ],d[],f[]; int main() { int…

题目http://poj.org/problem?id=3628 分析:给定一堆牛的高度,把牛叠加起来的高度超过牛棚的高度. 且是牛叠加的高度与牛棚高度之差最小. 把牛叠加的高度看作是背包的容量,利用01背包计算所能达到的最大值. 然后在最大值里面选择一个与牛棚高度差值最小的. 在开辟dp[]数组的时候没有必要开辟20*1000000不然会超过内存,适当小一点即可. #include<stdio.h>#include<string.h>const int INF=0XFFFFFF;…

Halloween treats Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 7644   Accepted: 2798   Special Judge Description Every year there is the same problem at Halloween: Each neighbour is only willing to give a certain total number of sweets…

Find a multiple Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 7192   Accepted: 3138   Special Judge Description The input contains N natural (i.e. positive integer) numbers ( N <= 10000 ). Each of that numbers is not greater than 15000…

The Pilots Brothers' refrigerator Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 22286   Accepted: 8603   Special Judge Description The game “The Pilots Brothers: following the stripy elephant” has a quest where a player needs to open a…

Flip Game Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 37427   Accepted: 16288 Description Flip game is played on a rectangular 4x4 field with two-sided pieces placed on each of its 16 squares. One side of each piece is white and the…

Corn Fields Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 9806   Accepted: 5185 Description Farmer John has purchased a lush new rectangular pasture composed of M by N (1 ≤ M ≤ 12; 1 ≤ N ≤ 12) square parcels. He wants to grow some yumm…

Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050   Accepted: 10989 Description Some positive integers can be represented by a sum of one or more consecutive prime numbers. How many such representatio…

Tree Recovery Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 11939   Accepted: 7493 Description Little Valentine liked playing with binary trees very much. Her favorite game was constructing randomly looking binary trees with capital le…

Seek the Name, Seek the Fame Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 17898   Accepted: 9197 Description The little cat is so famous, that many couples tramp over hill and dale to Byteland, and asked the little cat to give names t…

题目: poj 2352 Stars 数星星 题意:已知n个星星的坐标.每个星星都有一个等级,数值等于坐标系内纵坐标和横坐标皆不大于它的星星的个数.星星的坐标按照纵坐标从小到大的顺序给出,纵坐标相同时则按照横坐标从小到大输出. (0 <= x, y <= 32000) 要求输出等级0到n-1之间各等级的星星个数. 分析: 这道题不难想到n平方的算法,即从纵坐标最小的开始搜,每次找它前面横坐标的值比它小的点的个数,两个for循环搞定,但是会超时. 所以需要用一些数据结构去优化,主要是优化找 横坐…

poj   1251  Jungle Roads  (最小生成树) Link: http://poj.org/problem?id=1251 Jungle Roads Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 23507   Accepted: 11012 Description The Head Elder of the tropical island of Lagrishan has a problem. A b…