POJ 3624【水01背包】
原题链接:http://poj.org/problem?id=3624
我的链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=20437#problem/A欢迎来水
Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 14963 | Accepted: 6840 |
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 from the N (1 ≤ N≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).
Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di
Output
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
Sample Input
4 6 1 4 2 6 3 12 2 7
Sample Output
23
Source
题意:第一行的两个整数n 和 m 分别表示可选择放入的物品有n个,背包体积为m,
接下来 n行每行的前一个整数代表当前物品放入背包的体积和产生的价值。
每种物品只可以放一次,求放入物品总体积不超过背包总体积情况下可以产生的最大
价值。
算法:裸01背包。。。一切重头开始吧
状态转移方程:
for i=1..N
ZeroOnePack(c[i],w[i]);
for v=V..cost
f[v]=max{f[v],f[v-cost]+weight}
//Accepted 216 KB 297 ms C++ 343 B 2013-03-13 22:00:10 #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn = 12880 + 10; int dp[maxn]; int main() { int n,m; int w,d; scanf("%d%d", &n, &m); for(int i =1; i <= n; i++) { scanf("%d%d", &w, &d); for(int j = m; j >=w; j--) { dp[j] = max(dp[j],dp[j-w]+d); } } printf("%d\n", dp[m]); return 0; }