hdu 1114(完全背包)

时间：2016-04-15 20:20:00 阅读：248 评论：0 收藏：0 [点我收藏+]

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Piggy-Bank

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18924 Accepted Submission(s): 9579

Problem Description

Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it‘s weight in grams.

Output

Print exactly one line of output for each test case. The line must contain the sentence "The minimum amount of money in the piggy-bank is X." where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line "This is impossible.".

Sample Input

Sample Output

The minimum amount of money in the piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible.

题意：一个存钱罐有两个重量(单位为克),一个空的时候的重量，一个满的时候的重量，给定n组硬币,每组硬币都有它的价值和重量,每种硬币数量无穷多,求当前重量的存钱罐里面的钱最少为多少.
分析：每种物品无限多,可以分析出来是个完全背包问题.dp[i]代表钱罐重量为i时能够装的最少价值。
　　状态转移方程 dp[i] = min(dp[i],dp[i-V[i]]+W[i]) 其中V[i]是正向枚举的,因为每种物品无限多.

///分析：每种物品无限多,可以分析出来是个完全背包问题.dp[i]代表钱罐重量为i时能够装的最少价值。
///状态转移方程 dp[i] = min(dp[i],dp[i-V[i]]+W[i]) 其中V[i]是正向枚举的,因为每种物品无限多.
#include<stdio.h>
#include<algorithm>
#include<string.h>
#include<iostream>
#include <math.h>
#define N 10005
using namespace std;

int dp[N];
int V[505],W[505];
int INF = 99999999;
int main()
{
    int tcase ;
    scanf("%d",&tcase);
    while(tcase--){
        int E,F,n;
        scanf("%d%d",&E,&F);
        F-=E;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d%d",&W[i],&V[i]);
        }
        for(int i=1;i<=F;i++) dp[i] = INF;
        dp[0] = 0;  ///初始化重量为0的背包能够装的最大价值为0
        for(int i=1;i<=n;i++){
            for(int v=V[i];v<=F;v++){
                dp[v] = min(dp[v],dp[v-V[i]]+W[i]);
            }
        }
        if(dp[F]<INF)
        printf("The minimum amount of money in the piggy-bank is %d.\n",dp[F]);
        else printf("This is impossible.\n");
    }
    return 0;
}

hdu 1114(完全背包)

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原文地址：http://www.cnblogs.com/liyinggang/p/5396680.html

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