HDU 2126 Buy the souvenirs
来源:互联网 发布:python 哈希表 编辑:程序博客网 时间:2024/04/30 05:11
Buy the souvenirs
Time Limit: 10000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1283 Accepted Submission(s): 454
Problem Description
When the winter holiday comes, a lot of people will have a trip. Generally, there are a lot of souvenirs to sell, and sometimes the travelers will buy some ones with pleasure. Not only can they give the souvenirs to their friends and families as gifts, but also can the souvenirs leave them good recollections. All in all, the prices of souvenirs are not very dear, and the souvenirs are also very lovable and interesting. But the money the people have is under the control. They can’t buy a lot, but only a few. So after they admire all the souvenirs, they decide to buy some ones, and they have many combinations to select, but there are no two ones with the same kind in any combination. Now there is a blank written by the names and prices of the souvenirs, as a top coder all around the world, you should calculate how many selections you have, and any selection owns the most kinds of different souvenirs. For instance:
And you have only 7 RMB, this time you can select any combination with 3 kinds of souvenirs at most, so the selections of 3 kinds of souvenirs are ABC (6), ABD (7). But if you have 8 RMB, the selections with the most kinds of souvenirs are ABC (6), ABD (7), ACD (8), and if you have 10 RMB, there is only one selection with the most kinds of souvenirs to you: ABCD (10).
And you have only 7 RMB, this time you can select any combination with 3 kinds of souvenirs at most, so the selections of 3 kinds of souvenirs are ABC (6), ABD (7). But if you have 8 RMB, the selections with the most kinds of souvenirs are ABC (6), ABD (7), ACD (8), and if you have 10 RMB, there is only one selection with the most kinds of souvenirs to you: ABCD (10).
Input
For the first line, there is a T means the number cases, then T cases follow.
In each case, in the first line there are two integer n and m, n is the number of the souvenirs and m is the money you have. The second line contains n integers; each integer describes a kind of souvenir.
All the numbers and results are in the range of 32-signed integer, and 0<=m<=500, 0<n<=30, t<=500, and the prices are all positive integers. There is a blank line between two cases.
In each case, in the first line there are two integer n and m, n is the number of the souvenirs and m is the money you have. The second line contains n integers; each integer describes a kind of souvenir.
All the numbers and results are in the range of 32-signed integer, and 0<=m<=500, 0<n<=30, t<=500, and the prices are all positive integers. There is a blank line between two cases.
Output
If you can buy some souvenirs, you should print the result with the same formation as “You have S selection(s) to buy with K kind(s) of souvenirs”, where the K means the most kinds of souvenirs you can buy, and S means the numbers of the combinations you can buy with the K kinds of souvenirs combination. But sometimes you can buy nothing, so you must print the result “Sorry, you can't buy anything.”
Sample Input
24 71 2 3 44 01 2 3 4
Sample Output
You have 2 selection(s) to buy with 3 kind(s) of souvenirs.Sorry, you can't buy anything.
题目大意:
告诉你有几种纪念品以及分别对应的价格,要你用手中的钱去买纪念品,然后求出最多能买多少种纪念品以及最优方案数目。
大致思路:
这是一道01背包题目。要求最多能买多少种不难,难的是求出最优方案一共有多少种组合。
至于如何求最优方案数目,请看此文:
http://blog.csdn.net/mengt2012/article/details/43491859
知道如何求解最优方案,基本代码就出来了。一开始写的代码老是WA…后来才知道fill(kry, kry+maxm+1, INF)越界了…
代码如下:
//2126.cpp -- Buy the souvenirs#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <iomanip>typedef long long ll;using namespace std;const int maxn = 30 + 10;const int maxm = 500 + 10;const int INF = 1;int n, m;int w[maxn], dp[maxm], kry[maxm];void solve(){ fill(kry, kry+maxm, INF ); memset(dp, 0, sizeof(dp)); for( int i=1; i<=n; i++ ) { for( int j=m; j>=w[i]; j-- ) { if( dp[j]<dp[j-w[i]]+1) { dp[j] = dp[j-w[i]] + 1; kry[j] = kry[j-w[i]]; } else if( dp[j]==dp[j-w[i]]+1) kry[j] = kry[j] + kry[j-w[i]]; } } /* for(int j=0; j<maxm; j++ )kry[j] = 1; memset(dp, 0, sizeof(dp)); for( int i=1; i<=n; i++ ) { for( int j=1; j<=m; j++ ) { dp[i][j] = dp[i-1][j]; kry[i][j] = kry[i-1][j]; if( j>=w[i] ) { if( dp[i][j]<dp[i-1][j-w[i]]+1 ) { dp[i][j] = dp[i-1][j-w[i]] + 1; kry[i][j] = kry[i-1][j-w[i]]; } else if( dp[i][j]==dp[i-1][j-w[i]]+1 ) kry[i][j] = kry[i-1][j] + kry[i-1][j-w[i]]; } } } */ if( dp[m]!=0 ) printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n", kry[m], dp[m]); else printf("Sorry, you can't buy anything.\n");}int main(void){ int t; cin>>t; while( t-- ) { cin>>n>>m; for( int i=1; i<=n; i++ ) scanf("%d", &w[i]); solve(); } return 0;}
0 0
- hdu 2126 Buy the souvenirs
- hdu 2126 Buy the souvenirs
- Buy the souvenirs(hdu(2126)
- HDU 2126 Buy the souvenirs
- HDU 2126 Buy the souvenirs
- HDU 2126 Buy the souvenirs
- hdu 2126 Buy the souvenirs
- HDU - 2126 Buy the souvenirs
- hdu 2126 Buy the souvenirs
- Buy the souvenirs - HDU 2126 背包dp
- HDU 2126 - Buy the souvenirs(01背包)
- HDU-2126 Buy the souvenirs (DP)
- hdu 2126 Buy the souvenirs 01背包
- 2126 Buy the souvenirs
- HDU 2126 Buy the souvenirs (dp 二维01背包)
- hdu 2126 Buy the souvenirs(二维0/1背包)
- HDU 2126 Buy the souvenirs(DP:01背包)
- hdu 2126 Buy the souvenirs(求方案数)
- Android 屏幕适配
- 感恩
- poj1523 割点+连通分量
- iOS 7 SDK:后台传输服务
- 民事诉讼中的证据!
- HDU 2126 Buy the souvenirs
- [解题报告]多项式求导
- 哈希(hash)算法
- 初学abench压力测试
- IOS8 设置TableView Separatorinset 分割线从边框顶端开始
- 外挂辅助编写-分析动作数组与攻击捡物功能
- Linux下Maven的安装与使用
- 筛选法找出X~Y范围内的素数
- ava.lang.RuntimeException: Unable to instantiate activity ComponentInfo{*****Activity}: java.lang.Cl