DP:Space Elevator(POJ 2392)

题目大意:一群牛想造电梯到太空,电梯都是由一个一个块组成的,每一种块不能超过这个类型的高度,且每一种块都有各自的高度,有固定数量,问最高能造多高。
这题就是1742的翻版,对ai排个序就可以了
(尼玛,我qsort排了n-1个数,wa半天不知所措)
#include <iostream>
#include <functional>
#include <algorithm> using namespace std;
typedef struct _set
{
int h_i;
int max_h;
int count;
}Block;
int fcomp(const void *a, const void *b)
{
return (*(Block *)a).max_h - (*(Block *)b).max_h;
} static int dp[];
static Block B_Set[]; void Search(const int); int main(void)
{
int n;
while (~scanf("%d", &n))
{
if (n == ) continue;
for (int i = ; i <= n; i++)
scanf("%d%d%d", &B_Set[i].h_i, &B_Set[i].max_h, &B_Set[i].count);
qsort(B_Set, n + , sizeof(Block), fcomp);
Search(n);
}
return ;
} void Search(const int n)
{
int i, j;
memset(dp, -, sizeof(dp));
dp[] = ;
for (i = ; i <= n; i++)
{
if (B_Set[i].h_i == ) continue;
for (j = ; j < B_Set[i].h_i && j <= B_Set[i].max_h; j++)//先把前面的几个包确定下来
if (dp[j] != -)
dp[j] = B_Set[i].count;
for (; j <= B_Set[i].max_h; j++)
{
if (dp[j] == -)
{
if (dp[j - B_Set[i].h_i] <= )
continue;
else dp[j] = dp[j - B_Set[i].h_i] - ;
}
else dp[j] = B_Set[i].count;
}
}
for (int i = B_Set[n].max_h; i >= ; i--)
{
if (dp[i] >-)
{
printf("%d\n", i);
break;
}
}
}

DP:Space Elevator(POJ 2392)的更多相关文章
- POJ 2392 Space Elevator(贪心+多重背包)
POJ 2392 Space Elevator(贪心+多重背包) http://poj.org/problem?id=2392 题意: 题意:给定n种积木.每种积木都有一个高度h[i],一个数量num ...
- poj 2392 Space Elevator(多重背包+先排序)
Description The cows are going to space! They plan to achieve orbit by building a sort of space elev ...
- POJ 2392 Space Elevator(多重背包变形)
Q: 额外添加了最大高度限制, 需要根据 alt 对数据进行预处理么? A: 是的, 需要根据 alt 对数组排序 Description The cows are going to space! T ...
- poj[2392]space elevator
Description The cows are going to space! They plan to achieve orbit by building a sort of space elev ...
- A - Space Elevator(动态规划专项)
A - Space Elevator Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u ...
- poj2392 Space Elevator(多重背包问题)
Space Elevator Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 8569 Accepted: 4052 ...
- POJ2392:Space Elevator
Space Elevator Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 9244 Accepted: 4388 De ...
- BZOJ 1739: [Usaco2005 mar]Space Elevator 太空电梯
题目 1739: [Usaco2005 mar]Space Elevator 太空电梯 Time Limit: 5 Sec Memory Limit: 64 MB Description The c ...
- Building a Space Station POJ - 2031
Building a Space Station POJ - 2031 You are a member of the space station engineering team, and are ...
随机推荐
- CODEVS 1258 关路灯
写动归终于能不看题解一次A了!(其实交了两次,一次80一次A) 我练功发自真心! 题目描述 Description 多瑞卡得到了一份有趣而高薪的工作.每天早晨他必须关掉他所在村庄的街灯.所有的街灯都被 ...
- POJ1860Currency Exchange(Bellman + 正权回路)
Currency Exchange Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 23938 Accepted: 867 ...
- --------------------------PHP中访问数据库时带来的响应速度的问题解决-------------------------------------
----------------------------------------------------------------上图是秒,下图是毫秒比较TTFB-------------------- ...
- js中初学函数的使用
<script> function SetColor(name,value) { var oDiv=document.getElementById('div3'); oDiv.style[ ...
- Eigenvectors and eigenvalues
http://setosa.io/ev/eigenvectors-and-eigenvalues/ Explained Visually Tweet By Victor Powell and Lew ...
- 解决:[INS-20802] Oracle Net Configuration Assistant failed
在linux 中安装Oracle 11G 的时辰呈现 [INS-20802] Oracle Net Configuration Assistant failed 是oracle数据库的鼓掌,须要补丁p ...
- html5视频播放器
<!doctype html> <html> <head> <meta http-equiv="Content-Type" content ...
- String与Date、Timestamp互转
一.String与Date(java.util.Date)互转 1.1 String -> Date String dateStr = "2010/05/04 12:34:23&quo ...
- struts2 + ajax + json的结合使用,实例讲解
struts2用response怎么将json值返回到页面javascript解析,这里介绍一个struts2与json整合后包的用法. 1.准备工作 ①ajax使用Jquery:jquery-1.4 ...
- 调用MySql 分页存储过程带有输入输出参数
Create PROCEDURE getuser ( IN pageIndex INT, IN pageSize INT, OUT count INT ) BEGIN )*pageSize; sele ...