记录路径的背包问题

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我是靠谱客的博主缓慢母鸡，这篇文章主要介绍记录路径的背包问题，现在分享给大家，希望可以做个参考。

H - 多重背包记录方案

Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

Charlie is a driver of Advanced Cargo Movement, Ltd. Charlie drives a lot and so he often buys coffee at coffee vending machines at motorests. Charlie hates change. That is basically the setup of your next task.

Your program will be given numbers and types of coins Charlie has and the coffee price. The coffee vending machines accept coins of values 1, 5, 10, and 25 cents. The program should output which coins Charlie has to use paying the coffee so that he uses as many coins as possible. Because Charlie really does not want any change back he wants to pay the price exactly.

Input

Each line of the input contains five integer numbers separated by a single space describing one situation to solve. The first integer on the line P, 1 <= P <= 10 000, is the coffee price in cents. Next four integers, C1, C2, C3, C4, 0 <= Ci <= 10 000, are the numbers of cents, nickels (5 cents), dimes (10 cents), and quarters (25 cents) in Charlie's valet. The last line of the input contains five zeros and no output should be generated for it.

Output

For each situation, your program should output one line containing the string "Throw in T1 cents, T2 nickels, T3 dimes, and T4 quarters.", where T1, T2, T3, T4 are the numbers of coins of appropriate values Charlie should use to pay the coffee while using as many coins as possible. In the case Charlie does not possess enough change to pay the price of the coffee exactly, your program should output "Charlie cannot buy coffee.".

Sample Input

复制代码

Sample Output

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1
2
Throw in 2 cents, 2 nickels, 0 dimes, and 0 quarters.
Charlie cannot buy coffee.

主要还是借助増维度的方式来记录数据.....然后滚动即可！！！

感谢羊大神！！！！！！！！

复制代码

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#include<cstdio>
#include<cstring>
struct noi{
int val;
int amo;
};
int main()
{
int coffee,i,j,T[10000][5]={0};
noi coin[4];
while(scanf("%d%d%d%d%d",&coffee,&coin[0].amo,&coin[1].amo,&coin[2].amo,&coin[3].amo)!=EOF){
if(coffee==0&&coin[0].amo==0&&coin[1].amo==0&&coin[2].amo==0&&coin[3].amo==0)break;
memset(T,0,sizeof(T));
coin[0].val=1,coin[1].val=5,coin[2].val=10,coin[3].val=25;
for(i=0;i<4;i++){
for(j=coin[i].val;j<=coin[i].val*coin[i].amo&&j<=coffee;j++){
//
printf("%d %d %d %dn",i+1,j,T[j][0],T[j-coin[i].val][0]+1);
if(T[j][0]<T[j-coin[i].val][0]+1){
T[j][0]=T[j-coin[i].val][0]+1;
T[j][1]=T[j-coin[i].val][1];
T[j][2]=T[j-coin[i].val][2];
T[j][3]=T[j-coin[i].val][3];
T[j][4]=T[j-coin[i].val][4];
T[j][i+1]++;
}
}
}
// printf("total?%dn",T[coffee][0]);
if(T[coffee][0]==0)printf("Charlie cannot buy coffee.n");
else printf("Throw in %d cents, %d nickels, %d dimes, and %d quarters.n",T[coffee][1],T[coffee][2],T[coffee][3],T[coffee][4]);
}
return 0;
}