POJ 1384 Piggy-Bank(完全背包)
Description
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
Output
Sample Input
3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4
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.
思路:
1. 完全背包, 且要求恰好装满, dp[0] = 0
2. dp[i][j] 表示前 i 件物品放入容量为 j 的背包的最小价值, dp[i][j] = min(dp[i-1][j-k*w[i]]+k*v[i]])
3. 求 min, 要求初始化为 INF
代码:
#include <iostream>
using namespace std; const int INF = 0X3F3F3F3F;
int E, F, N;
int v[510], w[510];
int dp[10010]; int solve_dp() {
memset(dp, 0x3f, sizeof(dp));
dp[0] = 0;
int V = F-E;
for(int i = 0; i < N; i++) {
for(int j = w[i]; j <= V; j++) {
dp[j] = min(dp[j], dp[j-w[i]]+v[i]);
}
}
return dp[V];
} int main() {
freopen("E:\\Copy\\ACM\\测试用例\\in.txt", "r", stdin);
int tc;
cin >> tc;
while(tc--) {
scanf("%d%d%d", &E, &F, &N);
for(int i = 0; i < N; i++)
scanf("%d%d", &v[i], &w[i]);
// mainfunc
int ans = solve_dp();
if(ans == INF)
printf("This is impossible.\n");
else
printf("The minimum amount of money in the piggy-bank is %d.\n", ans);
}
return 0;
}
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