Proud Merchants(01背包变形)hdu3466】的更多相关文章

Proud Merchants Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1978    Accepted Submission(s): 792 Problem Description Recently, iSea went to an ancient country. For such a long time, it was…
Proud Merchants Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) Total Submission(s) : 12   Accepted Submission(s) : 5 Problem Description Recently, iSea went to an ancient country. For such a long time, it was the mo…
题目链接: 传送门 Proud Merchants Time Limit: 1000MS     Memory Limit: 65536K Description Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are…
题目链接:Knapsack problem 大意:给出T组测试数据,每组给出n个物品和最大容量w.然后依次给出n个物品的价值和体积. 问,最多能盛的物品价值和是多少? 思路:01背包变形,因为w太大,转而以v为下标,求出价值对应的最小体积,然后求出能够满足给出体积的最大价值. 经典题目,思路倒是挺简单的,就是初始化总觉得别扭...T_T大概,因为我要找的是最小值,所以初始化为maxn,就结了? 这个问题好像叫01背包的超大背包... 模拟一下样例吧! 1 5 15 // 初始化为dp[0] =…
01背包变形,注意dp过程的时候就需要取膜,否则会出错. 代码如下: #include<iostream> #include<cstdio> #include<cstring> using namespace std; #define MAXW 15005 #define N 155 #define LL long long #define MOD 1000000007 int w1[N],w2[N]; LL dp1[MAXW],dp2[MAXW]; int main(…
Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4739    Accepted Submission(s): 2470 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took pa…
http://acm.hdu.edu.cn/showproblem.php?pid=2955 [题意] 有一个强盗要去几个银行偷盗,他既想多抢点钱,又想尽量不被抓到.已知各个银行 的金钱数和被抓的概率,以及强盗能容忍的最大被抓概率.求他最多能偷到多少钱? [思路] 01背包:每个物品代价是每个银行钱的数目,物品的价值是在该银行不被抓的概率 (1-被抓概率),背包容量是所有银行钱的总和.01背包求dp[i]表示获得i的钱不被抓的最大概率.最后从大到小枚举出 dp[i]>=(1-P)这个i就是答案了…
C. Dima and Salad 题意 有n种水果,第i个水果有一个美味度ai和能量值bi,现在要选择部分水果做沙拉,假如此时选择了m个水果,要保证\(\frac{\sum_{i=1}^ma_i}{\sum_{i=1}^mb_i}==k\),问沙拉最大的美味度是多少? 思路 01背包变形. 对于给出的公式,我们化简一下: \(\sum_{i=1}^ma_i-k*\sum_{i=1}^mb_i==0\) 就变成了把a[i]-k*b[i]作为体积,a[i]作为价值,向容量为0的背包里放,可以取得的…
I - Proud Merchants Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u Description Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the…
需要排序的01背包. 这种题排序时只需要考虑两个怎么排,重载小于号就可以了. 需要注意的是,如果一个物品你想先放进背包里,那么你排序是要放到后面!01背包的放置顺序的倒着的! 看到别人的博客都只是比较了q-p,表示不解,但是都能AC...是谁错了呢?或者我这种表达式可以化成那个样子? AC代码: #include <iostream> #include <cstring> #include <algorithm> using namespace std; // (1 ≤…