POJ 1276 Cash Machine -- 动态规划(背包问题)
题目地址:http://poj.org/problem?id=1276
Description
Dk the machine has a supply of nk bills. For example,
N=3, n1=10, D1=100, n2=4, D2=50, n3=5, D3=10
means the machine has a supply of 10 bills of @100 each, 4 bills of @50 each, and 5 bills of @10 each.
Call cash the requested amount of cash the machine should deliver and write a program that computes the maximum amount of cash less than or equal to cash that can be effectively delivered according to the available bill supply of the machine.
Notes:
@ is the symbol of the currency delivered by the machine. For instance, @ may stand for dollar, euro, pound etc.
Input
cash N n1 D1 n2 D2 ... nN DN
where 0 <= cash <= 100000 is the amount of cash requested, 0 <=N <= 10 is the number of bill denominations and 0 <= nk <= 1000 is the number of available bills for the Dk denomination, 1 <= Dk <= 1000, k=1,N. White spaces can occur freely between the numbers
in the input. The input data are correct.
Output
Sample Input
735 3 4 125 6 5 3 350
633 4 500 30 6 100 1 5 0 1
735 0
0 3 10 100 10 50 10 10
Sample Output
735
630
0
0
#include <stdio.h>
#include <string.h> #define MAX 100001
#define MAXN 11
#define Max(a, b) (a) > (b) ? (a) : (b) int cash;
int num[MAXN];
int deno[MAXN];
int dp[MAX]; void ZeroOnePack (int deno){
int i;
for (i=cash; i>=deno; --i)
dp[i] = Max(dp[i], dp[i-deno] + deno);
} void CompletePack (int deno){
int i;
for (i=deno; i<=cash; ++i)
dp[i] = Max(dp[i], dp[i-deno] + deno);
} void MultiplePack (int deno, int num){
if (deno * num >= cash)
CompletePack (deno);
else{
int k = 1;
while (k < num){
ZeroOnePack (deno * k);
num -= k;
k *= 2;
}
ZeroOnePack (deno * num);
}
} int main(void){
int N;
int i; while (scanf ("%d%d", &cash, &N) != EOF){
for (i=1; i<=N; ++i){
scanf ("%d%d", &num[i], &deno[i]);
}
memset (dp, 0, sizeof(dp));
for (i=1; i<=N; ++i){
MultiplePack (deno[i], num[i]);
} printf ("%d\n", dp[cash]);
} return 0;
}
参考资料:背包问题九讲
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