题意:给定一个n*m的网格,每个格子里有A矿和B矿数量,A必须由右向左运,B只能从下向上运,中间不能间断,问最大总数量. 析:一个简单DP,dp[i][j] 表示 从 (0, 0) 到 (i, j) 最大人运输量.要么向左运输,要么向上运输,取最大值. 代码如下: #pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include…

分析: 对于网格grid[i][j]如果放向上的管道,那么grid[i][k], k>j 就只能放向上的管道了. 那么定义dp[i][j]表示第i行,最后一个放向左的管道是j的最大总矿量. j = 0表示全放向上,j = m表示全放向左. 如果grid[i][j]要往放向上的管道的话,前提是grid[i-1][j]也要是向上的管道. 因此转移为dp[i][j] = max(dp[i-1][k]+sumA(1,j)+sumB(j+1,n) ), k >= j 对于N= 500这样的规模,转移有点…

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. 思路 :记忆化搜索.也可以用递推. #include <stdio.h> #include <string.h> #include <iostream> using namespace std ; ][] ,blog[][] ; ][] ; int DP(int row…

Description The NASA Space Center, Houston, is less than 200 miles from San Antonio, Texas (the site of the ACM Finals this year). This is the place where the astronauts are trained for Mission Seven Dwarfs, the next giant leap in space exploration.…

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,首先我题目看了老半天,看懂后写出前缀和又不知道该如何定义状态好,写不出状态转移方程,太弱了. dp[i][j]表示以(i, j)为右下角时求得的最大值,状态转移方程:dp[i][j] = max (dp[i-1][j] + sum1[i][j], dp[i][j-1] + sum2[i][j]); sum1表示列的…

Martian Mining Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 2251 Accepted: 1367 Description The NASA Space Center, Houston, is less than 200 miles from San Antonio, Texas (the site of the ACM Finals this year). This is the place where t…

Martian Mining Time Limit:3000MS     Memory Limit:0KB     64bit IO Format:%lld & %llu Submit Status Practice UVA 1366 #include <stdio.h> #include <string.h> #include <algorithm> using namespace std; ][][]; int main() { int n,m; ][],s…

Martian Mining Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 2194   Accepted: 1326 Description The NASA Space Center, Houston, is less than 200 miles from San Antonio, Texas (the site of the ACM Finals this year). This is the place whe…

Super Jumping! Jumping! Jumping! Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 33384    Accepted Submission(s): 15093 Problem Description Nowadays, a kind of chess game called “Super Jumping!…

题目链接:http://codeforces.com/problemset/problem/455/A 给你n个数,要是其中取一个大小为x的数,那x+1和x-1都不能取了,问你最后取完最大的和是多少. 简单dp,dp[i]表示取i时zui最大和为多少,方程为dp[i] = max(dp[i - 1] , dp[i - 2] + cont[i]*i). #include <bits/stdc++.h> using namespace std; typedef __int64 LL; ; LL a…