Piggy-Bank - 九度教程第 102 题

Piggy-Bank - 九度教程第 102 题

题目

时间限制：1 秒 内存限制：32 兆 特殊判题：否
题目描述：
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：
3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 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.

题目大意：有一个储蓄罐，告知其空时的重量和当前重量，并给定一些钱币的价值和相应的重量，求储蓄罐中最少有多少现金。

由于每个钱币的数量都可以有任意多，所以该问题为完全背包问题。但是在该题中，完全背包有两处变化：
首先，要求的不再是最大值，而变为了最小值，这就要求在状态转移时，在 dp[j]和 dp[j-list[i].w]+list[i].v 中选择较小的转移值；
其次，该问题要求钱币和空储蓄罐的重量恰好达到总重量，即在背包问题中表现为背包恰好装满。
在前文中已经讨论了 0-1 背包的此类变化，只需变化 dp[j]的初始值即可。

#include <stdio.h>
#define INF 0x7fffffff

int min(int a,int b){
    return a<b ? a : b;
}

struct E{//代表钱币结构体
int w;
int v;
}list[501];

int dp[10001];//状态

int main()
{
    int T;
    scanf("%d",&T);//输入测试数据组数
    while(T--){//T次循环，处理T组数据
        int s,tmp;
        scanf("%d%d",&tmp,&s);//输入空储蓄罐数量和装满钱币的储蓄罐重量
        s-=tmp;//计算钱币所占重量
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d%d",&list[i].v,&list[i].w);
        }

        for(int i=0;i<=s;i++){
            dp[i]=INF;
        }

        dp[0]=0;
        //因为要求所有物品恰好装满，所以初始时除dp[0]外
        //其余dp[i]均为无穷（或者不存在）

        for(int i=1;i<=n;i++){//遍历所有物品
            for(int j=list[i].w;j<=s;j++){
                //完全背包，顺序遍历所有可能转移的状态
                if(dp[j-list[i].w]!=INF){
                    //若dp[j-list[i].w]不为无穷，就可以由此状态转移而来
                    dp[j]=min(dp[j],dp[j-list[i].w]+list[i].v);
                    //取转移值和原值的较小值
                }
            }
        }

        if(dp[s]!=INF){
            //若存在一种方案使背包恰好装满，输出其最小值
            printf("The mininum amount of money in the piggy-bank is %d.\n",dp[s]);
        }else{
            puts("This is impossible.");
        }
    }
    return 0;
}

总结一下完全背包问题：
其特点为每个物品可选的数量为无穷,其解法与 0-1背包整体保持一致，与其不同的仅为状态更新时的遍历顺序。时间复杂度和空间复杂度均和 0-1 背包保持一致。

最后介绍多重背包问题，其介于 0-1 背包和完全背包之间：有容积为V 的背包，给定一些物品，每种物品包含体积 w、价值 v、和数量 k，求用该背包能装下的最大价值总量。

与之前的背包问题都不同，每种物品可选的数量不再为无穷或者 1，而是介于其中的一个确定的数 k。与之前讨论的问题一样，可以将多重背包问题直接转化到 0-1 背包上去，即每种物品均被视为 k 种不同物品，对所有的物品求 0-1背包，其时间复杂度为：

由此可见，降低每种物品的数量 ki 将会大大降低其复杂度，于是采用一种更为有技巧性的拆分。将原数量为 k 的物品拆分为若干组，每组物品看成一件物品，其价值和重量为该组中所有物品的价值重量总和，每组物品包含的原物品个数分别为：为：1、2、4…k-2^c+1，其中 c 为使 k-2^c+1 大于 0 的最大整数。这种类似于二进制的拆分，不仅将物品数量大大降低，同时通过对这些若干个原物品组合得到新物品的不同组合，可以得到 0 到 k 之间的任意件物品的价值重量和，所以对所有这些新物品做 0-1 背包，即可得到多重背包的解。由于转化后的 0-1 背包物品数量大大降低，其时间复杂度也得到较大优化，为：