To My Girlfriend Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 1326    Accepted Submission(s): 515 Problem Description Dear GuoI never forget the moment I met with you.You carefully asked me:…

题目链接: http://acm.hdu.edu.cn/showproblem.php?pid=5800 To My Girlfriend Time Limit: 2000/2000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others) 问题描述 Dear Guo I never forget the moment I met with you.You carefully asked me: "I have a very diffic…

传送门:hdu 5800 To My Girlfriend 题意:给定n个物品,其中i,j必选,l,m必不选,问组成体积为s的方法一共有多少种 思路:定义dp[i][j][s1][s2],表示前i种物品能够构成的体积为j,其中有s1种定为必选,s2种定为不必选;因为递推到第i层时,只与第i-1层有关,所以把第一维降到2来省内存.然后就是dp[i][j][s1][s2]=dp[i-1][j][s1][s2]+dp[i-1][j][s1][s2-1]+dp[i-1][j-a[i]][s1-1][s2…

背包变形.dp[i][j][g][h]表示前i个数字,和为j,有g个必选,有h个必不选的方案数. 答案为sum{dp[n][j][2][2]}*4 #pragma comment(linker, "/STACK:1024000000,1024000000") #include<cstdio> #include<cstring> #include<cmath> #include<algorithm> #include<vector&g…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5534 题意: 给你度为1 ~ n - 1节点的权值,让你构造一棵树,使其权值和最大. 思路: 一棵树上每个节点的度至少为1,且度的和为2*n - 2.那么我们先给这些节点的度都-1,剩下的节点度为n - 2.此时我们发现,任意分配剩下的这些度给节点,都可以形成一棵树.这就变成了一个完全背包题,容量为n-2.注意dp要初始化为-inf.思路确实巧妙. #include <iostream> #inc…

[题目链接]http://acm.hdu.edu.cn/showproblem.php?pid=5800 [题目大意] 给出一个容量上限s,f[i][j][k][l][m]表示k和l两个物品不能选,i和j两个物品必选,最终质量为m的方案数.求这些方案数的总和. [题解] 令dp[i][j][s1][s2]表示前i个物品填了j的体积,有s1个物品选为为必选,s2个物品选为必不选的方案数(0<=s1,s2<=2),则有转移方程dp[i][j][s1][s2]=dp[i-1][j][s1][s2]+…

Proud Merchants Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1978    Accepted Submission(s): 792 Problem Description Recently, iSea went to an ancient country. For such a long time, it was…

题意: 有 n 个团队和 m 艘船,每艘船的载客量为 k,每个团队的人数为ai+1 ,转载该团队可获利润 bi,要求每个团队的所有人必须在同一艘船上, 且团队优先级高的团队所在船编号不能大于优先级低的团队,求可以获得的最大利润. 题解:其实没什么,只需要01背包就可以了,只不过优先考虑团队优先级高的.分析:dp[i] 表示获得 i 利润时需要的最少船位,且要保证优先级高的团队优先考虑. #include<cstdio> #include<cmath> #include<ios…

题目大意:中文题就不多说了 题目思路:由题意可知,只要高于5元,就可以随便刷,那我们就把最贵的留在最后刷.但是如果低于5元就什么也不能刷(哪怕你要买的物品价格不足五元),所以我们可以先求出(n-5)元的情况下最多能花掉多少钱,最后再减去最贵的物品价格就可以了,具体看代码吧. #include<cstdio> #include<stdio.h> #include<cstdlib> #include<cmath> #include<iostream>…

Bone Collector II Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4739    Accepted Submission(s): 2470 Problem Description The title of this problem is familiar,isn't it?yeah,if you had took pa…