poj1742 Coins 动态规划多重背包待补完

最新推荐文章于 2020-05-19 12:34:42 发布

kyoma

最新推荐文章于 2020-05-19 12:34:42 发布

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分类专栏： ------------解题报告------------- ------动态规划-------- 多重背包待补完

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------动态规划--------

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解决一个经典的计算机科学问题，即如何使用有限数量的特定面额硬币来支付从1到m的所有价格，而不找零。该问题通过动态规划算法求解。

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Coins

Time Limit: 3000MS		Memory Limit: 30000K
Total Submissions: 35350		Accepted: 12015

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box 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 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

8
4

Source

LouTiancheng@POJ

详细教学内容请参照挑战程序设计竞赛p62～p64

糊里糊涂地A了

为什么在j<a[i]的时候对dp的更新赋值是-1，还是没有很明白，回宿舍的路上在自习思考一下

想清楚了再补完这篇

代码：

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <stack>
using namespace std;
const int maxn = 1e5 + 10;
int val[maxn];
int num[maxn];
int dp[maxn];
int main(){
    int n,m,i,j;
    while(scanf("%d%d",&n,&m)){
        if(n==0&&m==0){
            break;
        }
        for(i=1;i<=n;i++){
            scanf("%d",val+i);
        }
        for(i=1;i<=n;i++){
            scanf("%d",num+i);
        }
        memset(dp, -1, sizeof(dp));
        dp[0]=0;
        for(i=1;i<=n;i++){
            for(j=0;j<=m;j++){
                if(dp[j]>=0){
                    dp[j] = num[i];
                }else if(j<val[i]||dp[j-val[i]]<=0){
                    dp[j] = -1;
                }else{
                    dp[j] = dp[j-val[i]] - 1;
                }
            }
        }
        int res = 0;
        for(i=1;i<=m;i++){
            if(dp[i]>=0){
                res++;
            }
        }
        printf("%d\n",res);
    }
    
    return 0;
}