POJ 3040 Allowance(贪心，诠释思想的好题)

Allowance

Time Limit: 1000MS

Memory Limit: 65536K

Total Submissions: 2300

Accepted: 945

Description

As a reward for record milk production, Farmer John has decided to start paying Bessie the cow a small weekly allowance. FJ has a set of coins in N (1 <= N <= 20) different denominations, where each denomination of coin evenly divides the next-larger denomination (e.g., 1 cent coins, 5 cent coins, 10 cent coins, and 50 cent coins).Using the given set of coins, he would like to pay Bessie at least some given amount of money C (1 <= C <= 100,000,000) every week.Please help him ompute the maximum number of weeks he can pay Bessie.

Input

* Line 1: Two space-separated integers: N and C 

* Lines 2..N+1: Each line corresponds to a denomination of coin and contains two integers: the value V (1 <= V <= 100,000,000) of the denomination, and the number of coins B (1 <= B <= 1,000,000) of this denomation in Farmer John's possession.

Output

* Line 1: A single integer that is the number of weeks Farmer John can pay Bessie at least C allowance

Sample Input

3 610 11 1005 120

Sample Output

111

Hint

INPUT DETAILS: 
FJ would like to pay Bessie 6 cents per week. He has 100 1-cent coins,120 5-cent coins, and 1 10-cent coin. 

OUTPUT DETAILS: 
FJ can overpay Bessie with the one 10-cent coin for 1 week, then pay Bessie two 5-cent coins for 10 weeks and then pay Bessie one 1-cent coin and one 5-cent coin for 100 weeks.

题意：约翰要给他的牛贝西发工资，每天不得低于C元，约翰有n种面值的钱币，第i种的面值为v_i，数量有b_i。问这些钱最多给贝西发多少天的工资。注意，每种面值的金钱都是下一种的面值的倍数。

题解：好题，深入了解贪心思想。POJ真不错啊，刷个贪心都把我搞得不要不要的(；′⌒`)。这一题分三步解决：

     1. 按照面值从大到小取，面值大于等于C的，直接取光。

     2. 再按面值从大到小取，凑近C，可以小于等于C，但不能大于C。

    

     3.最后从小到大取，凑满C，这里的凑满可以等于大于C。然后将上述2,3步取到的面值全部取走，再转入步骤2，这样每次找到的取法就是当前最优取法，直到所剩下的金币总价值不够C结束。

很好的题目，很能表现贪心思想：从局部的最优解找出总体最优解。

代码如下：

#include<cstdio>#include<cstring>#include<algorithm>#define INF 0x3f3f3f using namespace std;int use[30];//记录当前取法的第i种面值取的个数 struct node{int v,b;}a[25];int cmp(node a,node b){return a.v<b.v;} int main(){int n,c,i,cnt,ans,k,m;while(scanf("%d%d",&n,&c)!=EOF){for(i=0;i<n;++i)scanf("%d%d",&a[i].v,&a[i].b);sort(a,a+n,cmp);ans=0;for(i=n-1;i>=0;i--)//第一步，满足大于C的面值全部取走 {if(a[i].v>=c){ans+=a[i].b;a[i].b=0;}}while(1)//每次循环都在找一次当前最优取法，直到剩下的总金额小于C元 {int sign=0;cnt=c;memset(use,0,sizeof(use));for(i=n-1;i>=0;--i)//第二步，从大到小取，不能超过C的值 {if(a[i].b){k=cnt/a[i].v;m=min(k,a[i].b);cnt-=m*a[i].v;use[i]=m;if(cnt==0){sign=1;break;}}}if(cnt>0){for(i=0;i<n;++i)//第三步，从小到大取，凑满C {if(a[i].b>use[i]){while(use[i]<a[i].b){cnt-=a[i].v;use[i]++;if(cnt<=0){sign=1;break;}}}if(sign)break;}}if(!sign)break;m=INF;for(i=0;i<n;++i){if(use[i])//找到当前取法的能取的总次数 m=min(m,a[i].b/use[i]);}ans+=m;for(i=0;i<n;++i){if(use[i])a[i].b-=m*use[i];} }printf("%d\n",ans);}return 0;}