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…

http://acm.sdut.edu.cn:8080/vjudge/contest/view.action?cid=259#problem/H 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 across the globe. Nowadays,th…

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…

 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…

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…

  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 destro…

HDOJ(HDU).3466 Dividing coins ( DP 01背包 无后效性的理解) 题意分析 要先排序,在做01背包,否则不满足无后效性,为什么呢? 等我理解了再补上. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 505 #define nn 505*100 using namespace std;…

HDOJ(HDU).2546 饭卡(DP 01背包) 题意分析 首先要对钱数小于5的时候特别处理,直接输出0.若钱数大于5,所有菜按价格排序,背包容量为钱数-5,对除去价格最贵的所有菜做01背包.因为这时候还剩下5块钱,直接买最贵的那个菜,就可以保证剩下来的钱数是最小的. 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nma…

HDOJ(HDU).2602 Bone Collector (DP 01背包) 题意分析 01背包的裸题 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 1005 using namespace std; int v[nmax],w[nmax],dp[nmax]; int main() { //freopen("in…

In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4764    Accepted Submission(s): 1569 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): 5220    Accepted Submission(s): 1745 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): 5472    Accepted Submission(s): 1843 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project t…

10397507 2014-03-25 23:30:21 Accepted 1203 0MS 480K 428 B C++ 泽泽 题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1203 Problem Description Speakless很早就想出国,现在他已经考完了所有需要的考试,准备了所有要准备的材料,于是,便需要去申请学校了.要申请国外的任何大学,你都要交纳一定的申请费用,这可是很惊人的.Speakless没有多少钱,总共只攒了n万美元…

Problem Description 电子科大本部食堂的饭卡有一种非常诡异的设计,即在购买之前推断剩余金额. 假设购买一个商品之前,卡上的剩余金额大于或等于5元,就一定能够购买成功(即使购买后卡上剩余金额为负),否则无法购买(即使金额足够).所以大家都希望尽量使卡上的剩余金额最少. 某天,食堂中有n种菜出售.每种菜可购买一次. 已知每种菜的价格以及卡上的剩余金额,问最少可使卡上的剩余金额为多少.   Input 多组数据.对于每组数据: 第一行为正整数n.表示菜的数量.n<=1000. 第二行…

In Action Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4859    Accepted Submission(s): 1600 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project…

zhx and contest Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 455    Accepted Submission(s): 158 Problem Description As one of the most powerful brushes in the world, zhx usually takes part i…

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…

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…

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…

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> 题目大意: 给出一个0-n组成的图,1-n的点上分布着值为pow的电站,给出图的m条边以及距离,从0出发到n个点中的x个点的行走距离和最小(因为是每炸一个点派出一辆坦克),且x个点的pow之和必须超过总的pow和的一半. 解题分析: 由于本题数据范围很小,只有100,所以我们能够用floyed算出0到任意一点的最短距离,然后将所有0可达的点看成物品,0到它们的最短距离看成体积,这样将所有可达物品最短距离之和看成背包容量,这些可达点的pow看成价值,然后再用01背包.dp[i]表示…

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…

In Action Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4720 Accepted Submission(s): 1553 Problem Description Since 1945, when the first nuclear bomb was exploded by the Manhattan Project team i…

/* 题意:有 n 个站点(编号1...n),每一个站点都有一个能量值,为了不让这些能量值连接起来,要用 坦克占领这个站点!已知站点的 之间的距离,每个坦克从0点出发到某一个站点,1 unit distance costs 1 unit oil! 最后占领的所有的站点的能量值之和为总能量值的一半还要多,问最少耗油多少! */ /* 思路:不同的坦克会占领不同的站点,耗油最少那就是路程最少,所以我们先将从 0点到其他各点的 最短距离求出来!也就是d[i]的值!然后我们又知道每一个站点的所具有的能量…

三国志 时间限制:3000 ms  |  内存限制:65535 KB 难度:5   描述 <三国志>是一款很经典的经营策略类游戏.我们的小白同学是这款游戏的忠实玩家.现在他把游戏简化一下,地图上只有他一方势力,现在他只有一个城池,而他周边有一些无人占的空城,但是这些空城中有很多不同数量的同种财宝.我们的小白同学虎视眈眈的看着这些城池中的财宝. 按照游戏的规则,他只要指派一名武将攻占这座城池,里面的财宝就归他所有了.不过一量攻占这座城池,我们的武将就要留守,不能撤回.因为我们的小白手下有无数的武…

意甲冠军:给出的数量和袋骨骼的数,然后给每块骨骼的价格值和音量.寻求袋最多可容纳骨骼价格值 难度;这个问题是最基本的01背包称号,不知道的话,推荐看<背包9说话> AC by SWS 主题链接 http://acm.hdu.edu.cn/showproblem.php?pid=2602 代码: #include<stdio.h> #include<string.h> typedef struct{ int w, v; }str; str s[1005]; int dp[…

题目传送门 题意:有n个金矿,每个金矿有抓取的消耗的时间和价值,矿工在原点,问在T时间内能得到的最大的价值 分析:唯一和01背包不同的是金矿可能共线,也就是抓取近的金矿后才能抓后面共线的金矿.这是分组背包问题,方法是将点按照斜率排序,如果相等按照距离原点远近排序,将斜率相等的点分成一组,每组的点累加上前面的点的时间和价值,这样每组只选一个点,就是01背包了 收获:分组背包问题 代码: /************************************************ * Auth…

01背包,求第k大. 以前看k短路的时候看过代码以为懂了 = =结果还是跑去看了别人的代码才会.果然要自己写一遍才行啊 0.0难得1A.. 每次把可能的2k种求出来,求前k个.注意要不一样的k个数.. /************************************************************** Problem : 2639 ( Bone Collector II ) Judge Status : Accepted RunId : 15212165 Languag…

描述 <三国志>是一款很经典的经营策略类游戏.我们的小白同学是这款游戏的忠实玩家.现在他把游戏简化一下,地图上只有他一方势力,现在他只有一个城池,而他周边有一些无人占的空城,但是这些空城中有很多不同数量的同种财宝.我们的小白同学虎视眈眈的看着这些城池中的财宝. 按照游戏的规则,他只要指派一名武将攻占这座城池,里面的财宝就归他所有了.不过一量攻占这座城池,我们的武将就要留守,不能撤回.因为我们的小白手下有无数的武将,所以他不在乎这些. 从小白的城池派出的武将,每走一公理的距离就要消耗一石的粮食,…

http://acm.sdut.edu.cn:8080/vjudge/contest/view.action?cid=187#problem/G 该题要注意的就是每张单子A种类的总和不能大与600,同样B,C类也一样,还有注意如果不是A,B,C类的不可以报销: 该题就是要把浮点型变成整数这样才能用01背包,这里就只要乘以100就可以了. 这题考的背包很简单,就是输入的金额为背包的容积,债券既是物体的体积又是物体的利润.就是处理输入的数据有点麻烦,这是我所不擅长的. #include <iostr…