【POJ 3265】Problem Solving

Problem Solving

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 1645		Accepted: 675

Description

In easier times, Farmer John's cows had no problems. These days, though, they have problems, lots of problems; they have P (1 ≤ P ≤ 300) problems, to be exact. They have quit providing milk and have taken regular jobs like all other good citizens. In fact, on a normal month they make M (1 ≤ M ≤ 1000) money.

Their problems, however, are so complex they must hire consultants to solve them. Consultants are not free, but they are competent: consultants can solve any problem in a single month. Each consultant demands two payments: one in advance (1 ≤ payment ≤ M) to be paid at the start of the month problem-solving is commenced and one more payment at the start of the month after the problem is solved (1 ≤payment ≤ M). Thus, each month the cows can spend the money earned during the previous month to pay for consultants. Cows are spendthrifts: they can never save any money from month-to-month; money not used is wasted on cow candy.

Since the problems to be solved depend on each other, they must be solved mostly in order. For example, problem 3 must be solved before problem 4 or during the same month as problem 4.

Determine the number of months it takes to solve all of the cows' problems and pay for the solutions.

Input

Line 1: Two space-separated integers: M and P. 
Lines 2..P+1: Line i+1 describes problem i with two space-separated integers: B_i and A_i. B_i is the payment to the consult BEFORE the problem is solved; A_i is the payment to the consult AFTER the problem is solved. 

Output

Line 1: The number of months it takes to solve and pay for all the cows' problems.

Sample Input

100 5
40 20
60 20
30 50
30 50
40 40

Sample Output

6

Hint

+------+-------+--------+---------+---------+--------+
|      | Avail | Probs  | Before  | After   | Candy  |
|Month | Money | Solved | Payment | Payment | Money  |
+------+-------+--------+---------+---------+--------+
| 1    | 0     | -none- | 0       | 0       | 0      |
| 2    | 100   | 1, 2   | 40+60   | 0       | 0      |
| 3    | 100   | 3, 4   | 30+30   | 20+20   | 0      |
| 4    | 100   | -none- | 0       | 50+50   | 0      |
| 5    | 100   | 5      | 40      | 0       | 60     |
| 6    | 100   | -none- | 0       | 40      | 60     | 
+------+-------+--------+---------+---------+--------+

Source

USACO 2007 January Gold

 

一道简单的动态规划题目。

设f[n]为答案。

枚举j，表示j~i这几个问题在同一个月内完成（如果可以），那么很明显有两种情况，做完j-1个问题的最小月数是f[j-1]，下个月要付before[j-1]的钱。

　　（1）before[j-1] + sigma{a[k] | j <= k <= i} <= M,那么这样的最小值就是,f[j - 1]+1了；

　　（2）before[j-1] + sigma{a[k] | j <= k <= i} > M,那么很明显不能和上面的问题直接接上去了，f[j-1]+2；

时间复杂度O(N²)。

#include<cstdio>
#include<cstring>
#include<algorithm>

using namespace std;

int m, p, a[305], b[305];
int f[305], before[305];

int main()
{
    scanf("%d%d", &m, &p);
    for (int i = 1; i <= p; ++i)
        scanf("%d%d", &a[i], &b[i]);
    before[0] = a[0] = 300005;
    f[0] = 1;
    for (int i = 1; i <= p; ++i)
    {
        int suma = 0, sumb = 0;
        f[i] = f[i - 1] + 2;
        before[i] = b[i];
        for (int j = i; j >= 1; --j)
        {
            suma += a[j];
            sumb += b[j];
            if (suma > m || sumb > m) break;
            if (suma + before[j - 1] <= m && f[j - 1] + 1 <= f[i])
            {
                if (f[j - 1] + 1 == f[i]) before[i] = min(before[i], sumb);
                else before[i] = sumb;
                f[i] = f[j - 1] + 1;
            }
            if (suma + before[j - 1] > m && f[j - 1] + 2 <= f[i])
            {
                if (f[j - 1] + 2 == f[i]) before[i] = min(before[i], sumb);
                else before[i] = sumb;
                f[i] = f[j - 1] + 2;
            }
        }
    }
    printf("%d
", f[p]);
    return 0;
}