首先我是按这篇文章来确定题目的. poj3624 Charm Bracelet 模板题 没有要求填满,所以初始化为0就行 #include<cstdio> #include<algorithm> #include<cmath> #include<iostream> #include<cstring> using namespace std; ]; ]; int n,m; +]; int main() { // freopen("inpu…

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 public that they are both smart and fun. In order to do this,…

题目链接:http://poj.org/problem?id=2184 Cow Exhibition Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 9479   Accepted: 3653 Description "Fat and docile, big and dumb, they look so stupid, they aren't much  fun..."  - Cows with Guns by…

Charm Bracelet Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 38909   Accepted: 16862 Description Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible fro…

题目链接:http://poj.org/problem?id=3624 Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list…

http://poj.org/problem?id=3624 题意:给出物品的重量和价值,在重量一定的情况下价值尽可能的大. 思路:经典0-1背包.直接套用模板. #include<iostream> #include<algorithm> using namespace std; ; int n, m; int dp[maxn]; ], D[]; int main() { //freopen("D:\\txt.txt", "r", stdi…

题目链接: http://poj.org/problem?id=2151 Check the difficulty of problems Time Limit: 2000MSMemory Limit: 65536K 问题描述 Organizing a programming contest is not an easy job. To avoid making the problems too difficult, the organizer usually expect the contes…

题目链接: http://poj.org/problem?id=1112 Team Them Up! Time Limit: 1000MSMemory Limit: 10000K 问题描述 Your task is to divide a number of persons into two teams, in such a way, that: everyone belongs to one of the teams; every team has at least one member; e…

POJ.3624 Charm Bracelet(DP 01背包) 题意分析 裸01背包 代码总览 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #define nmax 13000 #define nnmax 3500 using namespace std; int dp[nmax]; int w[nnmax],d[nnmax]; int main…

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…