POJ 3624 01背包】的更多相关文章

初学DP,用贪心的思想想解题,可是想了一个多小时还是想不出. //在max中的两个参数f[k], 和f[k-weight[i]]+value[i]都是表示在背包容量为k时的最大价值 //f[k]是这个意思,就不用说了. //而f[k-weight[i]]+value[i]也表示背包容量为k时的最大价值是为什么呢? //首先,f[k-weight[i]]表示的是背包容量为k-weight[i]的容量,也就是说f[k-weight[i]] //表示的是容量还差weiht[i]才到k的价值,+walu…
POJ 2184 Cow Exhibition Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 14657   Accepted: 5950 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…
Cow Exhibition Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 10200   Accepted: 3977 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…
Balance Time Limit: 1000 MS Memory Limit: 30000 KB 64-bit integer IO format: %I64d , %I64u Java class name: Main [Submit] [Status] [Discuss] Description Gigel has a strange "balance" and he wants to poise it. Actually, the device is different fr…
思路: 1.01背包 先找到所有奶牛身高和与B的差. 然后做一次01背包即可 01背包的容积和价格就是奶牛们身高. 最后差值一减输出结果就大功告成啦! 2. 搜索 这思路很明了吧... 搜索的确可以过- 3. 模拟! 0到1< < n 来一遍.(状压呗) 01背包的: // by SiriusRen #include <cstdio> #include <algorithm> using namespace std; int N,B,sum=0,h[25],f[1000…
Proud Merchants Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)Total Submission(s): 4257    Accepted Submission(s): 1757 Problem Description Recently, iSea went to an ancient country. For such a long time, it was…
poj3624 Charm Bracelet 模板题 没有要求填满,所以初始化为0就行 #include<cstdio> #include<iostream> using namespace std; #define N 15010 int n,m,v[N],c[N],f[N]; int main(){ scanf("%d%d",&n,&m); ;i<=n;i++) scanf("%d%d",&v[i],&…
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…