Colored Rectangles【简单DP】

题目描述：原题链接

You are given three multisets of pairs of colored sticks:

R pairs of red sticks, the first pair has length r1, the second pair has length r2, …, the R-th pair has length rR;
G pairs of green sticks, the first pair has length g1, the second pair has length g2, …, the G-th pair has length gG;
B pairs of blue sticks, the first pair has length b1, the second pair has length b2, …, the B-th pair has length bB;
You are constructing rectangles from these pairs of sticks with the following process:

take a pair of sticks of one color;
take a pair of sticks of another color different from the first one;
add the area of the resulting rectangle to the total area.
Thus, you get such rectangles that their opposite sides are the same color and their adjacent sides are not the same color.

Each pair of sticks can be used at most once, some pairs can be left unused. You are not allowed to split a pair into independent sticks.

What is the maximum area you can achieve?

Input
The first line contains three integers R, G, B (1≤R,G,B≤200) — the number of pairs of red sticks, the number of pairs of green sticks and the number of pairs of blue sticks.

The second line contains R integers r1,r2,…,rR (1≤ri≤2000) — the lengths of sticks in each pair of red sticks.

The third line contains G integers g1,g2,…,gG (1≤gi≤2000) — the lengths of sticks in each pair of green sticks.

The fourth line contains B integers b1,b2,…,bB (1≤bi≤2000) — the lengths of sticks in each pair of blue sticks.

Output
Print the maximum possible total area of the constructed rectangles.

Examples
input
1 1 1
3
5
4
output
20

input
2 1 3
9 5
1
2 8 5
output
99

input
10 1 1
11 7 20 15 19 14 2 4 13 14
8
11
output
372

思路：

先将每颜色的长度从大到小排序，后3重循环暴力枚举出答案。因为n只有200。

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;
typedef long long ll;
const int N=230;ll r[N],g[N],b[N];
ll dp[N][N][N];
bool cmp(int x,int y)
{return x>y;
}
int main()
{int R,G,B; scanf("%d %d %d",&R,&G,&B);for(int i=1;i<=R;i++) scanf("%lld",&r[i]);for(int i=1;i<=G;i++) scanf("%lld",&g[i]);for(int i=1;i<=B;i++) scanf("%lld",&b[i]);sort(r+1,r+R+1,cmp);sort(g+1,g+G+1,cmp);sort(b+1,b+B+1,cmp);ll res=0;for(int i=1;i<=R+1;i++){for(int j=1;j<=G+1;j++){for(int k=1;k<=B+1;k++){dp[i][j][k]=max(max(dp[i][j][k],dp[i][j-1][k-1]+g[j-1]*b[k-1]),max(dp[i-1][j-1][k]+r[i-1]*g[j-1],dp[i-1][j][k-1]+r[i-1]*b[k-1]));res=max(res,dp[i][j][k]);}}}printf("%lld\n",res);return 0;
}

Colored Rectangles【简单DP】相关推荐

D. Colored Rectangles[思维dp]
题意: 3种颜色木棍,每次从两种颜色木棍各选一对组成矩形.求最后所有矩形面积和最大值. #include <cstdio> #include <cstring> #includ ...
Codeforces 41D Pawn 简单dp
题目链接:点击打开链接给定n*m 的矩阵常数k 以下一个n*m的矩阵,每一个位置由 0-9的一个整数表示问: 从最后一行開始向上走到第一行使得路径上的和 % (k+1) == 0 每一个格子仅仅 ...
hdu2067 简单dp或者记忆化搜索
题意: 小兔的棋盘 Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Sub ...
POJ1088:滑雪（简单dp)
题目链接: http://poj.org/problem?id=1088 题目要求: 一个人可以从某个点滑向上下左右相邻四个点之一,当且仅当高度减小.求可以滑落的最长长度. 题目解析: 首先要先排一 ...
第三讲数学与简单DP【完结】
目录 1205. 买不到的数目 [数学结论题] 1211. 蚂蚁感冒 [模拟 / 推理] 1216. 饮料换购 [简单 / 模拟] 2. 01背包问题 [板子题] 1015. 摘花生 [简单DP] 8 ...
hdu 2881（简单dp）
题意:n*n的矩阵,里面有m个格子是有任务要去完成的,t,x,y表示要在第t秒到达(x,y)的格子完成任务,问你最多可以完成多少解题思路:简单dp,将时间排个序后就是LIS #include< ...
P1005 矩阵取数游戏（__int128模板/简单dp）
转跳P1005 题目描述帅帅经常跟同学玩一个矩阵取数游戏:对于一个给定的 n \times mn×m 的矩阵,矩阵中的每个元素 a_{i,j}a i,j 均为非负整数.游戏规则如下: 每次取数时 ...
最少拦截系统，简单dp，（学长说这是贪心？！。。。。。。也是醉了）
description 某国为了防御敌国的导弹袭击,发展出一种导弹拦截系统.但是这种导弹拦截系统有一个缺陷:虽然它的第一发炮弹能够到达任意的高度,但是以后每一发炮弹都不能超过前一发的高度.某天, ...
HDU 1158【简单dp】
题意:给你一个项目,需要几个月来完成买,同时也给你每个月最少需要的工人数.并且告诉你hiring,firing每个工人的钱数,以及每个月应付每个工人的工资.求项目完成时最小花费. 这是个简单dp,思路 ...
4.15 每周作业 —— 简单DP
免费馅饼 Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other) Total Submissi ...

Colored Rectangles【简单DP】

题目描述：原题链接

思路：

Colored Rectangles【简单DP】相关推荐

最新文章

热门文章