Poj 3040 Allowance【贪心模拟】

Allowance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2542		Accepted: 1038

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 6
10 1
1 100
5 120

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：

有个人想给他的奶牛发工资，每周都发金额至少为c的钱，现在他有一些面值的硬币，给出对应的数量，而且面值之间呈倍数关系，问这个人的钱最多能发多少周。

题解：

最开始的理解是，如果刚好可以凑成面值C，那么肯定这样的方式是最优的解法，但是就好像成了动态一直更新讨论的感觉了，好复杂......

后来发现，原来面值之间有倍数关系，情况就相对简单了，因为大的面值能由小面值拼凑出来，因此完全刻意优先使用大的面值的，但是处理起来还是很复杂........

后来参考了大神的思路，发现自己思维还是没打开.......

其实贪心的策略是一样的，但是大神使用了一个辅助数组来存放每次某个物品需要选多少，并且在动态更新选举的个数，自己咋就没想到呢！！！

后来研读了很久，终于敢下手写总结了.....

贪心策略：

1，从面值最大的开始选取，如果超过了目标值，先不选（其实可以先特判累加这些超出的个数，算是一种优化吧）

2，预选一些小于目标值的硬币，使得选取的面值小于目标值，这时候的值但是肯定是小于目标值而且是非常接近目标值的

3，从面值最小的开始选取，选择使得当前值大于目标值的最小面值的那个硬币

4，如果当前值无法大于目标值，说明无法满足条件，结束执行，否则使用当前策略直接发去一些周数的工资，并且更新数据

5，重复执行，直至第4条结束执行。

/* http://blog.csdn.net/liuke19950717 */
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
struct node
{
ll val,num;
}x[30];
int cmp(node a,node b)
{
return a.val<b.val;
}
ll cal(ll n,ll c)
{
ll num[30]={0},sum=c;//数组存放每个元素取多少个
for(ll i=n-1;i>=0&∑--i)//尝试进行选取
{
ll tp=min(sum/x[i].val,x[i].num);//某一个最多能选多少
num[i]=tp;//记录数量
sum-=tp*x[i].val;//需求量减小
}
if(sum>0)//如果没能正好凑够
{
for(ll i=0;i<n;++i)//从最小的遍历
{
if(x[i].num>num[i]&&x[i].val>=sum)//选还剩下的，能满足需求的
{
++num[i];//选上
sum=0;//选好了
break;
}
}
}
if(sum>0)//说明无法满足
{
return 0;
}
ll ans=0x3f3f3f3f;//初始化一个无穷大
for(ll i=0;i<n;++i)
{
if(num[i])//找到当前策略能发的周数
{
ans=min(ans,x[i].num/num[i]);
}
}
for(ll i=0;i<n;++i)
{
x[i].num-=ans*num[i];//减去发去的工资
}
return ans;
}
int main()
{
ll n,c;
while(~scanf("%lld%lld",&n,&c))
{
for(ll i=0;i<n;++i)
{
scanf("%lld%lld",&x[i].val,&x[i].num);
}
sort(x,x+n,cmp);
ll num=cal(n,c),ans=0;
while(num)//执行到无法满足
{
ans+=num;
num=cal(n,c);
}
printf("%lld\n",ans);
}
return 0;
}