ACM 背包问题 Piggy-Bank

以下讲解的是最简单的背包问题...

推荐一个“背包九讲“——

http://blog.sina.com.cn/s/blog_8cf6e8d90100zldn.html

题目形式比较单一，例”有N件物品和一个容量为V的背包。第i件物品的重量是w[i]，价值是v[i]。求解将哪些物品装入背包可使这些物品的重量总和不超过背包容量，且价值总和最大。”这样的形式。

每个物品无非是装入背包或者不装入背包，那么就一个一个物品陆续放入背包中。

即f[i][v]表示前i件物品恰放入一个容量为v的背包可以获得的最大价值。

则其状态转移方程便是：f[i][v]=max{f[i-1][v],f[i-1][v-c[i]]+w[i]}。 

TOJ 1835 Piggy-Bank

描述

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!

输入

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, P and W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it's weight in grams.

输出

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.".

样例输入

310 11021 130 5010 11021 150 301 6210 320 4

样例输出

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.

题目意思：（举例第一组数据）有一个存钱罐，不放钱的时候是10g，现在是110g.里面有两种货币。一块钱1g种和三十块钱50g。问里面至少有多少钱。如果重量不能完全达到，打印一行“This is impossible.”。

 这是最简单的完全背包问题，直接扔代码啦~

#include<stdio.h>#include<string.h>#include<algorithm>#include<iostream>using namespace std;const int INF=0x3f3f3f3f;//等于十进制1061109567//写9个9也行，就是要让它最大。 int v[550],w[550],dp[10010],n,Value;int main(){//背包问题     int T,E,F,i,j;    scanf("%d",&T);    while(T--)    {        scanf("%d%d",&E,&F);        Value=F-E;        scanf("%d",&n);        for(i=0;i<n;i++)           scanf("%d%d",&w[i],&v[i]);        for(i=0;i<=Value;i++)dp[i]=INF;        dp[0]=0;        for(i=0;i<n;i++)          for(j=v[i];j<=Value;j++)            if(dp[j-v[i]]<INF)              dp[j]=min(dp[j],dp[j-v[i]]+w[i]);        if(dp[Value]>=INF)printf("This is impossible.\n");        else printf("The minimum amount of money in the piggy-bank is %d.\n",dp[Value]);    }    return 0;}