A_完全背包
2014-07-18 11:53
176 查看
/* copyright: Grant Yuan algorithm: 完全背包 time : 2014.7.18 __________________________________________________________________________________________________ A - 完全背包 基础 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit Status Description 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! Input 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, 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. Output 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.". Sample Input 3 10 110 2 1 1 30 50 10 110 2 1 1 50 30 1 6 2 10 3 20 4 */ #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<functional> #include<queue> #include<stack> #include<cstdlib> #define INF 999999999; using namespace std; long long w[501]; long long p[501]; long long dp[2][10001]; long long w1,w2; long long t,n; long long w3; int main() { cin>>t; while(t--){ cin>>w1>>w2; w3=w2-w1; cin>>n; for(int i=0;i<n;i++) cin>>p[i]>>w[i]; for(int i=0;i<2;i++) for(int j=1;j<10001;j++) dp[i][j]=INF; for(int i=0;i<n;i++) for(int j=0;j<=w3;j++) { if(j<w[i]) dp[(i+1)&1][j]=dp[i&1][j]; else dp[(i+1)&1][j]=min(dp[i&1][j],dp[(i+1)&1][j-w[i]]+p[i]); } if(dp[n&1][w3]<999999999) printf("The minimum amount of money in the piggy-bank is %lld.\n",dp[n&1][w3]); else printf("This is impossible.\n"); } return 0; }
相关文章推荐
- lightoj1200 【完全背包】
- DP总结(1) 01背包 完全背包 多重背包
- NYOJ 311 完全背包
- 背包问题——“完全背包”详解及实现(包含背包具体物品的求解)
- 【动态规划】完全背包问题
- POJ-2063 完全背包
- POJ 2063 Investment / 体积变大的完全背包
- C++背包01与完全背包
- uva242, 完全背包||记忆化搜索,(类似广工校赛的一道题)
- poj 1252 Euro Efficiency 完全背包,可以找零
- HDU 1284 钱币兑换问题 完全背包
- 01背包、完全背包、多重背包问题的C++实现及路径记录
- POJ 2063 Investment 变形的完全背包
- HDU 4508 湫湫系列故事——减肥记I(完全背包模板)
- 01背包、完全背包、多重背包归纳
- hihoCoder 1043 完全背包
- 【二维费用之完全背包】HDU3127-WHUgirls
- NYOJ311 完全背包
- codefroces 543A Writing Code dp优化 完全背包
- poj 1384Piggy-Bank(完全背包)