poj2184 Cow Exhibition

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POJ2184 Cow Exhibition[DP 状态负值]

Cow Exhibition Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 12420 Accepted: 4964 Description "Fat and docile, big and dumb, they look so stupid, they aren't much fun..." - Cows with Guns by Dana Lyons The cows want to prove to…

POJ-2184 Cow Exhibition(01背包变形)

Cow Exhibition Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 10949 Accepted: 4344 Description "Fat and docile, big and dumb, they look so stupid, they aren't much fun-" - Cows with Guns by Dana Lyons The cows want to prove to the pu…

poj2184 Cow Exhibition（p-01背包的灵活运用）

转载请注明出处:http://blog.csdn.net/u012860063 题目链接:id=2184">http://poj.org/problem?id=2184 Description "Fat and docile, big and dumb, they look so stupid, they aren't much fun..." - Cows with Guns by Dana Lyons The cows want to prove to the…

poj2184 Cow Exhibition【01背包】+【负数处理】+（求两个变量的和最大）

题目链接:https://vjudge.net/contest/103424#problem/G 题目大意: 给出N头牛,每头牛都有智力值和幽默感,然后,这个题目最奇葩的地方是,它们居然可以是负数!!现在叫你求出其中的牛中,智力值总数和幽默感总数加起来最大的值,当然,智力值的总数必须大于等于零,幽默感总数也是. 解题思路: 转载于>>>大牛博客对于每头牛,我们有两种选择,要么选,要么不选.看到这,就会觉得跟01背包很像,但是有 2 个变量来描述此状态的特征,智力,幽默感,并且他们可能是…

思路: dp+滚动数组. 类似01背包. 实现: #include <iostream> #include <cstdio> #include <algorithm> using namespace std; const int INF = 0x3f3f3f3f; ; ], b[], dp[][], n, p, q; int solve(int n) { ; i < ; i++) { for (int j = -t; j <= t; j++) { dp[i]…

POJ2184 Cow Exhibition 背包

题目大意:已知c[i]...c[n]及f[i]...f[n],现要选出一些i,使得当sum{c[i]}和sum{f[i]}均非负时,sum(c[i]+f[i])的最大值. 以sum(c[i])(c[i]>=0)作为背包容量totV,i当作物体,c[i]当作物体体积,f[i]当作物体重量,01背包后,求max{j+DP[j]} (非负)即可. 注意: 这里物体体积存在负数,所以循环j时起始条件不能为j=0!而应当在for下面加if,判断子问题在minJ~maxJ范围之内. 初值不是DP[最左端点]…

POJ 2184 Cow Exhibition【01背包+负数（经典）】

POJ-2184 [题意]: 有n头牛,每头牛有自己的聪明值和幽默值,选出几头牛使得选出牛的聪明值总和大于0.幽默值总和大于0,求聪明值和幽默值总和相加最大为多少. [分析]:变种的01背包,可以把幽默度看成体积,智商看成价值,那么就转换成求体积和价值都为正值的最大值的01背包了. 以 TS 作为体积,TF作为价值,在保证体积.价值非负的情况下,求解 sum,取其所有情况的最大值. 难点: 1)体积出现负数,将区间改变 [-100000, 100000] ---> [0, 200000]. (注…