In Action Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3339 Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project team in the US, the number of nuclear weapons have soared…

传送门: http://acm.hdu.edu.cn/showproblem.php?pid=3339 In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7869    Accepted Submission(s): 2674 Problem Description Since 1945, when the first…

题目链接:[http://acm.hdu.edu.cn/showproblem.php?pid=3339] In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5951    Accepted Submission(s): 1998 Problem Description Since 1945, when the firs…

In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3869    Accepted Submission(s): 1237 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project t…

In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5102    Accepted Submission(s): 1696 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project t…

题目链接: 题目 In Action Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) 问题描述 Since 1945, when the first nuclear bomb was exploded by the Manhattan Project team in the US, the number of nuclear weapons have soared across th…

 Since 1945, when the first nuclear bomb was exploded by the Manhattan Project team in the US, the number of nuclear weapons have soared across the globe. Nowadays,the crazy boy in FZU named AekdyCoin possesses some nuclear weapons and wanna destroy…

思路:最短路求出到每个点的最小代价,然后01背包,求出某一代价所能拿到的最大价值,然后搜索最后结果. 代码: #include<cstdio> #include<set> #include<cmath> #include<stack> #include<cstring> #include<string> #include<sstream> #include<iostream> #include<algor…

http://acm.hdu.edu.cn/showproblem.php?pid=3339 这道题就是dijkstra+01背包,先求一遍最短路,再用01背包求. #include <cstdio> #include <cstring> #include <algorithm> #define maxn 100000 using namespace std; <<; int n,m,st,ed,c; int dis[maxn]; bool vis[maxn…

分析:每种菜仅仅可以购买一次,但是低于5元不可消费,求剩余金额的最小值问题..其实也就是最接近5元(>=5)时, 购买还没有买过的蔡中最大值问题,当然还有一些临界情况 1.当余额充足时,可以随意购买菜,即∑p - max_p +5 <= m  时,re = m - ∑p 2.当余额不充足时,有一种特殊情况,不能消费的情况,即m<5时    re = m; 3.余额不足时,只能购买部分菜,转化成01背包问题,找出最接近最接近5的值, 状态转换方程: f[0][P] = true;f[0][…

HDU 3127 WHUgirls(完全背包) http://acm.hdu.edu.cn/showproblem.php? pid=3127 题意: 如今有一块X*Y的矩形布条, 然后有n种规格的x[i]*y[i]的小布条, 每种布条能够卖出val[i]的价值. 问你原始的X*Y布条最多能卖多少价值?   当中每次分割布条仅仅能水平或垂直的切, 且一刀究竟. 分析: 本题看起来比較麻烦, 可是搞懂原理后还是非常easy的. 把当前还剩余的矩形布条看成容量, 那么我们就是再这有限的容量里面要找出…

In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5220    Accepted Submission(s): 1745 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project t…

最短路+01背包 In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3857    Accepted Submission(s): 1229 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan…

http://acm.hdu.edu.cn/showproblem.php?pid=3535 题意:有n组工作,T时间,每个工作组中有m个工作,改组分类是s,s是0是组内至少要做一件,是1时最多做一件,2时随意,每项工作的描述是花费的时间和获得的快乐值,求在T时间内可获的最大快乐值. memset放错位置了,折腾老半天. 分组混合背包,有的取一件或不取,有的随意,有的最少一个 分三种情况讨论 s==0 考虑前面取过时这次取或不取,前一组取过时这次取或不取 s==1 考虑前一组取过时这次取或不取…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=4003 题目大意:有K个机器人,走完树上的全部路径,每条路径有个消费.对于一个点,机器人可以出去再回来,开销2倍.也可以不回来,一直停在某个点(如果你的机器人数量足够多的话).问最小开销. 解题思路: 其实这题只能说是类树形背包. 用dp[i][j]表示在i点,有j个不回来的机器人走过的最小开销. 比如dp[i][0]就表示,i点及其子点全部靠其它点的不回来的机器人探索.所以机器人是一来一回开销2倍…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1561 题目大意:从树根开始取点.最多取m个点,问最大价值. 解题思路: cost=1的树形背包. 有个虚根0,取这个虚根也要cost,所以最后的结果是dp[0][m+1]. 本题是cost=1的特殊背包问题,在两个for循环上有一个优化. for(f+1...j....cost) for(1....k...j-cost) 其中f为当前已经dfs子结点个数.之所以+1,是因为根要预留一个空间. f+…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=1011 题目大意:树上取点,先取父亲,再取儿子.每个点,权为w,花费为cost,给定m消费总额,求最大权和. 解题思路: 树形背包模板题.首先建一个无向图. 每个点的cost=(bug[root]+19)/20,即虫子数不满20也要派一个人. 用dp[i][j]表示以i为根的子树中,花费为j的最大权和. 转移方程:dp[i][j]=max(dp[i][j],dp[i][j-k]+dp[t][k]),…

AreYouBusy Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Problem Description Happy New Term!As having become a junior, xiaoA recognizes that there is not much time for her to AC problems, because there are some o…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1963 //多重背包 #include <cstdio> #include <cstring> #include <iostream> using namespace std; + ; #define N 15 long long dp[maxn], ans; int c[N], w[N], V; void Pack(int C, int W) { for(int i = C…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 题目分析:0-1背包  注意dp数组的清空, 二维转化为一维后的公式变化 /*Bone Collector Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 34192 Accepted Submission(s): 14066 Proble…

Partial Tree Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5534 Description In mathematics, and more specifically in graph theory, a tree is an undirected graph in which any two nodes are connected by exactly…

Piggy-Bank Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status Description Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action come…

Big Event in HDU Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 23728    Accepted Submission(s): 8363 Problem Description Nowadays, we all know that Computer College is the biggest department…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2602 这是做的第一道01背包的题目.题目的大意是有n个物品,体积为v的背包.不断的放入物品,当然物品有各自的体积和价值.在不超过总体积v的情况下,问能够达到的最大价值.并且物品是一个一个放入的.最后若有剩余的体积也不会填满. 刚开始是用贪心做的.将价值与体积的比值设定为一个值,即单位价值.然后按照单位价值排序,挨个取物品,考虑到了体积为0的情况,就将单位价值设定为无穷大.但是这样做并不能保证最优解.…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=2844 思路:多重背包 , dp[i] ,容量为i的背包最多能凑到多少容量,如果dp[i] = i,那么代表这个数能凑出来,ans+1: 实现代码: #include<bits/stdc++.h> using namespace std; ; int lis[M],dp[M],a[M],c[M]; int main() { int n,m,idx; while(cin>>n>>m…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=2955 思路:一开始看急了,以为概率是直接相加的,wa了无数发,这道题目给的是被抓的概率,我们应该先求出总的逃跑概率,1-逃跑概率就是最后被抓的概率,dp的话,以所有银行总金额为容量,以单个银行的金额为体积,以逃跑的概率为价值,跑01背包,最后找一下小于被抓概率的最大金额. 实现代码: #include<bits/stdc++.h> using namespace std; ; ],b[M]; int…

与 HDU 1028 相似的题目. 方法一:完全背包. 限制条件:硬币总值不超过 n. 目标:求出组合种数. 令 dp[ i ][ j ] == x 表示用前 i 种硬币组合价值为 j 的钱共 x 种方法. 状态转移方程:dp[ i ][ j ] = dp[ i - 1][ j ] + dp[ i ][ j - v[ i ] ] : 方程解释:用前 i 种硬币组合出钱 j 的方法数 = 前 i - 1 种硬币组合出钱 j 的方法数(不用第 i 种硬币)+ 至少用一枚第 i 种硬币的方法数. 滚动…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1114 思路:求出存钱罐装全部装满情况下硬币的最小数量,即求出硬币的最小价值.转换为最小背包的问题. #include<iostream> #include<cstdio> #include<cstring> using namespace std; ; int dp[INF],val[INF],we[INF]; int main(void) { int t,n,i,j,m,…

http://acm.hdu.edu.cn/showproblem.php?pid=5000 每只羊有n个属性 下面n个数字表示每个属性的值范围为[ 0, T[i] ] 对于羊圈里的a羊和b羊,若a羊的每个属性都>=b羊,则a羊会杀死b羊. 问羊圈里最多存活多少只羊. sum相同的羊不会互相杀死. sum不同的羊不会重合.(我们设a羊sum = x,b羊sum = y,若a,b羊能共存,但不会把ab同时放到羊圈里. 因为一定存在一只羊c ,sum = x,且c和b不能共存,既然不能共存,则我们放…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=2639 Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 5817    Accepted Submission(s): 3067 Problem Description The title of this…