(中等) POJ 2948 Martian Mining,DP。
Description
The Mars Odyssey orbiter identified a rectangular area on the surface of Mars that is rich in minerals. The area is divided into cells that form a matrix of n rows and m columns, where the rows go from east to west and the columns go from north to south. The orbiter determined the amount of yeyenum and bloggium in each cell. The astronauts will build a yeyenum refinement factory west of the rectangular area and a bloggium factory to the north. Your task is to design the conveyor belt system that will allow them to mine the largest amount of minerals.
There are two types of conveyor belts: the first moves minerals from east to west, the second moves minerals from south to north. In each cell you can build either type of conveyor belt, but you cannot build both of them in the same cell. If two conveyor belts of the same type are next to each other, then they can be connected. For example, the bloggium mined at a cell can be transported to the bloggium refinement factory via a series of south-north conveyor belts.
The minerals are very unstable, thus they have to be brought to the factories on a straight path without any turns. This means that if there is a south-north conveyor belt in a cell, but the cell north of it contains an east-west conveyor belt, then any mineral transported on the south-north conveyor beltwill be lost. The minerals mined in a particular cell have to be put on a conveyor belt immediately, in the same cell (thus they cannot start the transportation in an adjacent cell). Furthermore, any bloggium transported to the yeyenum refinement factory will be lost, and vice versa.
Your program has to design a conveyor belt system that maximizes the total amount of minerals mined,i.e., the sum of the amount of yeyenum transported to the yeyenum refinery and the amount of bloggium transported to the bloggium refinery.
题目就是给一个矩阵,然后问能够到最左边的所有直线和能够到最上边的所有直线经过的值的和。
DP求解,dp[i][j]表示第i列向右的矩形所形成的最小值,其中j表示第i列的最后以↑的位置,然后从右向左推就好。
代码如下:
// ━━━━━━神兽出没━━━━━━
// ┏┓ ┏┓
// ┏┛┻━━━━━━━┛┻┓
// ┃ ┃
// ┃ ━ ┃
// ████━████ ┃
// ┃ ┃
// ┃ ┻ ┃
// ┃ ┃
// ┗━┓ ┏━┛
// ┃ ┃
// ┃ ┃
// ┃ ┗━━━┓
// ┃ ┣┓
// ┃ ┏┛
// ┗┓┓┏━━━━━┳┓┏┛
// ┃┫┫ ┃┫┫
// ┗┻┛ ┗┻┛
//
// ━━━━━━感觉萌萌哒━━━━━━ // Author : WhyWhy
// Created Time : 2015年07月20日 星期一 10时39分58秒
// File Name : 2948.cpp #include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h> using namespace std; const int MaxN=; int N,M;
int C1[MaxN][MaxN];
int C2[MaxN][MaxN];
int dp[MaxN][MaxN]; int main()
{
//freopen("in.txt","r",stdin);
//freopen("out.txt","w",stdout); int maxn; while(~scanf("%d %d",&N,&M) && N+M)
{
for(int i=;i<=N;++i)
for(int j=;j<=M;++j)
scanf("%d",&C1[i][j]); for(int i=;i<=N;++i)
for(int j=;j<=M;++j)
scanf("%d",&C2[i][j]); memset(dp,,sizeof(dp)); for(int i=;i<=M;++i)
{
for(int j=;j<=N;++j)
C2[j][i]+=C2[j-][i]; for(int j=N-;j>=;--j)
C1[j][i]+=C1[j+][i];
} for(int i=M;i>=;--i)
for(int j=;j<=N;++j)
{
maxn=; for(int k=j;k<=N;++k)
maxn=max(maxn,dp[i+][k]); dp[i][j]=maxn+C2[j][i]+C1[j+][i];
} maxn=; for(int i=;i<=N;++i)
maxn=max(maxn,dp[][i]); printf("%d\n",maxn);
} return ;
}
(中等) POJ 2948 Martian Mining,DP。的更多相关文章
- POJ 2948 Martian Mining(DP)这是POJ第200道,居然没发现
题目链接 两种矿石,Y和B,Y只能从从右到左,B是从下到上,每个空格只能是上下或者左右,具体看图.求左端+上端最大值. 很容易发现如果想最优,分界线一定是不下降的,分界线上面全是往上,分界线下面都是往 ...
- poj 2948 Martian Mining (dp)
题目链接 完全自己想的,做了3个小时,刚开始一点思路没有,硬想了这么长时间,想了一个思路, 又修改了一下,提交本来没抱多大希望 居然1A了,感觉好激动..很高兴dp又有所长进. 题意: 一个row*c ...
- POJ 2948 Martian Mining
Martian Mining Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 2251 Accepted: 1367 Descri ...
- POJ 2948 Martian Mining(DP)
题目链接 题意 : n×m的矩阵,每个格子中有两种矿石,第一种矿石的的收集站在最北,第二种矿石的收集站在最西,需要在格子上安装南向北的或东向西的传送带,但是每个格子中只能装一种传送带,求最多能采多少矿 ...
- POJ 2498 Martian Mining
Martian Mining Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 2194 Accepted: 1326 De ...
- (中等) POJ 3034 Whac-a-Mole,DP。
Description While visiting a traveling fun fair you suddenly have an urge to break the high score in ...
- UVa 1366 - Martian Mining (dp)
本文出自 http://blog.csdn.net/shuangde800 题目链接: 点击打开链接 题目大意 给出n*m网格中每个格子的A矿和B矿数量,A矿必须由右向左运输,B矿必须由下向上运输 ...
- 递推DP UVA 1366 Martian Mining
题目传送门 /* 题意:抽象一点就是给两个矩阵,重叠的(就是两者选择其一),两种铺路:从右到左和从下到上,中途不能转弯, 到达边界后把沿途路上的权值相加求和使最大 DP:这是道递推题,首先我题目看了老 ...
- UVA 1366 九 Martian Mining
Martian Mining Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Sta ...
随机推荐
- iOS开发:后台运行以及保持程序在后台长时间运行
第一部分 1.先说说iOS 应用程序5个状态: 停止运行-应用程序已经终止,或者还未启动. 不活动-应用程序处于前台但不再接收事件(例如,用户在app处于活动时锁住了设备). 活动-app处于“使用中 ...
- 百度,google的地理编码
1.百度的地理编码:(不支持中国以外的其它城市) http://api.map.baidu.com/geocoder/v2/?ak=E974997f80db18330f8f5c61d084a677&a ...
- 设置TabBar分栏控制器上图片的大小问题
我们都知道,iOS因为屏幕分辨率的问题,UID在交付我们iOS开发人员程序配图的时候,一般是三套图,分别对应三种不同的分辨率,对不同size的屏幕系统会自动使用不同像素的图片,我们只需要在命名时给三套 ...
- 使用HttpUtils 上传视频文件
private void shangchuan(){ //文件的路径 //File file=new File(path); File fi ...
- jQuery 数据滚动(上下)
setInterval(function() { jq('.sjbg02 li:first').animate({ 'height': '0', 'opacity': '0' }, 'slow', f ...
- sql语句删除由于无主键导致完全重复的数据方法
sql语句删除由于无主键导致完全重复的数据方法 select distinct * into #Tmp from t_column drop table t_column select * into ...
- 一个不错的angular 字体库( 引用js文件就行)
https://klarsys.github.io/angular-material-icons/ 使用方法: 1.引入这个js文件, <script src="//cdnjs.clo ...
- VIEWCONTROLLER的启动流程
转载自:http://sunnyyoung.net/post/ios/2015-04-22-viewcontrollerde-qi-dong-liu-cheng-yu-jie-xi VIEWCONTR ...
- ms08_067利用过程
进入msf. show exploits. use exploit/windows/smb/ms08_067_netapi. show playloads. set PLAYLOAD windows/ ...
- DWR整合之JSF
DWR 与 JSF DWR 包括两个 JSF 的扩展点,一个创造器和一个 ServletFilter. 1.JSF Creator DWR1.1 中有一个体验版的 JsfCreator.你可以在 dw ...