Piggy-Bank
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6808 Accepted Submission(s): 3353
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!
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, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it's weight in grams.
cannot be reached exactly, print a line "This is impossible.".
3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4
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.比较典型的完全背包问题!!将d[0]初始化为0其他的都初始化为INF(因为求的是最小值,求最大值的话必须初始化为0)要求恰好装满的话只有d[0]的初值有合理解0;容量为零恰好装满的最优解自然为零,其他初始化为正无穷(因为他求最小值,即不存在合理解);#include<iostream> #include<stdlib.h> #include<math.h> #include<stdio.h> #include<algorithm> #include<string.h> #include<math.h> #include<stdlib.h> #include<vector> #define INF 1000000000 using namespace std; struct coin { int p; int w; }c[520]; int d[10000]; int main() { int t; scanf("%d",&t); for(int L = 0;L<t;++L) { for(int i = 1;i<10000;++i) d[i] = INF; d[0] = 0; int E,F; scanf("%d%d",&E,&F); int total = F - E; int n; scanf("%d",&n); for(int i=1;i<=n;++i) { scanf("%d%d",&c[i].p,&c[i].w); } for(int i=1;i<=n;++i) for(int j=c[i].w;j<=total;++j) { d[j] = min(d[j],d[j-c[i].w]+c[i].p); } if(d[total]==INF) printf("This is impossible.\n"); else printf("The minimum amount of money in the piggy-bank is %d.\n",d[total]); } return 0; }