HDU_多重背包系列

HDU_2191 悼念512汶川大地震遇难同胞——珍惜现在，感恩生活

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 24045    Accepted Submission(s): 10161

Problem Description

急！灾区的食物依然短缺！
为了挽救灾区同胞的生命，心系灾区同胞的你准备自己采购一些粮食支援灾区，现在假设你一共有资金n元，而市场有m种大米，每种大米都是袋装产品，其价格不等，并且只能整袋购买。
请问：你用有限的资金最多能采购多少公斤粮食呢？

后记：
人生是一个充满了变数的生命过程，天灾、人祸、病痛是我们生命历程中不可预知的威胁。
月有阴晴圆缺，人有旦夕祸福，未来对于我们而言是一个未知数。那么，我们要做的就应该是珍惜现在，感恩生活——
感谢父母，他们给予我们生命，抚养我们成人；
感谢老师，他们授给我们知识，教我们做人
感谢朋友，他们让我们感受到世界的温暖；
感谢对手，他们令我们不断进取、努力。 
同样，我们也要感谢痛苦与艰辛带给我们的财富～

 

Input

输入数据首先包含一个正整数C，表示有C组测试用例，每组测试用例的第一行是两个整数n和m(1<=n<=100, 1<=m<=100),分别表示经费的金额和大米的种类，然后是m行数据，每行包含3个数p，h和c(1<=p<=20,1<=h<=200,1<=c<=20)，分别表示每袋的价格、每袋的重量以及对应种类大米的袋数。

 

Output

对于每组测试数据，请输出能够购买大米的最多重量，你可以假设经费买不光所有的大米，并且经费你可以不用完。每个实例的输出占一行。

 

Sample Input

18 22 100 44 100 2

 

Sample Output

400

 

分析：多重背包，二进制优化，可做模板。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2191

代码清单：

#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 100 + 5;int C;int n, m;int p[maxn];int h[maxn];int c[maxn];int dp[maxn];void input() {cin >> n >> m;for(int i = 0; i < m; ++i) {cin >> p[i] >> h[i] >> c[i]; }}void solve() {memset(dp, 0, sizeof(dp));for(int i = 0; i < m; ++i) {if(p[i] * c[i] >= n) {for(int j = p[i]; j <= n; ++j) {dp[j] = max(dp[j - p[i]] + h[i], dp[j]);}}else {int k, ki;for(k = 0; ((1 << (k + 1)) - 1) < c[i]; ++k) {ki = (1 << k);for(int j = n; j >= p[i] * ki; --j) {dp[j] = max(dp[j - p[i] * ki] + h[i] * ki, dp[j]);}}ki = c[i] - ((1 << k) - 1);for(int j = n; j >= p[i] * ki; --j) {dp[j] = max(dp[j - p[i] * ki] + h[i] * ki, dp[j]);}}}cout << dp[n] << endl;}int main() {cin >> C;for(int t = 0; t < C; ++t) {input();solve();}return 0;}

HDU_2844 Coins

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 12164    Accepted Submission(s): 4852

Problem Description

Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse and found there were some coins. He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch.

You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

 

Input

The input contains several test cases. The first line of each test case contains two integers n(1 ≤ n ≤ 100),m(m ≤ 100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1 ≤ Ai ≤ 100000,1 ≤ Ci ≤ 1000). The last test case is followed by two zeros.

 

Output

For each test case output the answer on a single line.

 

Sample Input

3 101 2 4 2 1 12 51 4 2 10 0

 

Sample Output

84

 

题意：给出n个硬币的数量及价值，问在总价值限定为[1, m]的范围内，能够凑成多少种不同的价值。

分析：多重背包。把m当成容量，进行多重背包后，统计[1, m]内的价值有多少种。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=2844

代码清单：

#include <set>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 100 + 5;const int maxm = 100000 + 5;int n, m;int A[maxn];int C[maxn];int dp[maxm];set<int>cnt;void input() {for(int i = 0; i < n; ++i) {scanf("%d", &A[i]);}for(int i = 0; i < n; ++i) {scanf("%d", &C[i]);}}void solve() {memset(dp, 0, sizeof(dp));cnt.clear();for(int i = 0; i < n; ++i) {if(A[i] * C[i] >= m) {for(int j = A[i]; j <= m; ++j) {dp[j] = max(dp[j - A[i]] + A[i], dp[j]);}}else {int ki, k;for(k = 0; ((1 << (k + 1)) - 1) < C[i]; ++k) {ki = (1 << k);for(int j = m; j >= A[i] * ki; --j) {dp[j] = max(dp[j - A[i] * ki] + A[i] * ki, dp[j]);}}ki = C[i] - ((1 << k) - 1);for(int j = m; j >= A[i] * ki; --j) {dp[j] = max(dp[j - A[i] * ki] + A[i] * ki, dp[j]);}}}for(int i = 1; i <= m; ++i) {if(dp[i]) cnt.insert(dp[i]);}printf("%d\n", cnt.size());}int main() {while(scanf("%d%d", &n, &m) != EOF && n + m != 0) {input();solve();}return 0;}