POJ Problem 3040 Allowance 【贪心】

Allowance

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2932		Accepted: 1183

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.

Source

USACO 2005 October Silver

贪心思路：
    先将面值大于c发给Bessie的钱。
    每次从面值最大的开始，用need数组存放每个面值所需要的数目，并且该面值的使用总价值不大于c，
    要是不足用小于该面值的补, 但是不大于c。如果c不为零，从最小的面值开始加上，使得损失最小。
    如果还有剩余的，说明所剩的金额已经不足以支付下一个月的。
    每次使得损失最少，所以总的损失是最小的，所支付的周数最小。

#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX_N 105
#define MAX(a, b)   ((a > b)? a: b)
#define MIN(a, b)   ((a < b)? a: b)
using namespace std;

struct node {
    int val, num;
}mon[MAX_N];
bool cmp(node x, node y) {
    return x.val < y.val;
}
int need[MAX_N];
void init() {
    for (int i = 0; i < MAX_N; i++) {
        need[i] = 0;
    }
}
int main () {
    int c, n, m;
    while (scanf("%d%d", &n, &c) != EOF) {
        int ans = 0;
        for (int i = 0; i < n; i++) {
            scanf("%d%d", &mon[i].val, &mon[i].num);
        }
        int m = -1;
        sort(mon, mon + n, cmp);
        for (int i = n - 1; i >= 0; i--) {
            if (mon[i].val >= c) {
                ans += mon[i].num;
            }
            else {
                m = i;
                break;
            }
        }
        while (true) {
            init();
            int res = c;
            //先进行循环，从面值最大的开始求所需 各面值的数目。
            for (int i = m; i >= 0; i--) {
                if (mon[i].num && res) {
                    int k = res/mon[i].val;
                    k = MIN(k, mon[i].num);
                    need[i] = k;
                    res -= k*mon[i].val;
                }
            }
            //如果剩下的大于零，说明不能支付正好的金额，就从最小的面值选一张。
            if (res) {
                for (int i = 0; i <= m; i++) {
                    if (mon[i].val >= res && mon[i].num > need[i]) {
                        res = 0;
                        need[i]++;
                        break;
                    }
                }
            }
            //如果还有剩余，就结束。
            if(res) break;
            int week = 1e8;
            //找出可以支付的最小的周数
            for (int i = 0; i <= m; i++) {
                if (need[i])
                 week = MIN(week, mon[i].num/need[i]);
            }
            ans += week;
            for (int i = 0; i <= m; i++) {
                mon[i].num -= need[i]*week;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}
/*
5 1
9 6
9 1
6 2
4 0
1 8
34
*/