Problem 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: Bi and Ai. Bi is the payment to the consult BEFORE the problem is solved; Ai 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

题意：有p个问题，每人每月解决一个问题，解决问题有两个代价，当月给a[i]，第二月给b[i]，然后每个月钱有m，求最快多久能解决所有问题。

思路：

假设j之前为上个月及之前完成的，j到i为在本月完成的，那么：

当 before[j-1]+a[j...i]<=m 成立时，最小值为 f[j-1]+1

当 before[j-1]+a[j...i]>m 成立时，最小值为 f[j-1]+2

Source Program

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<string>
#include<cstdlib>
#include<queue>
#include<set>
#include<vector>
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define N 1001
#define MOD 123
#define E 1e-6
using namespace std;
int a[N],b[N];
int before[N],f[N];
int main()
{int m,p;scanf("%d%d",&m,&p);for(int i=1;i<=p;i++)scanf("%d%d",&a[i],&b[i]);before[0]=INF;f[0]=1;for(int i=1;i<=p;i++){int cost_a=0,cost_b=0;f[i]=f[i-1]+2;before[i]=b[i];for (int j=i;j>=1;j--)//j之前为上个月及之前完成的，j到i为在本月完成的{cost_a+=a[j];cost_b+=b[j];if (cost_a > m||cost_b>m)//超出当月钱数，退出break;if (cost_a+before[j-1]<=m&&f[j-1]+1<=f[i]){if (f[j-1]+1==f[i])before[i]=min(before[i],cost_b);elsebefore[i]=cost_b;f[i]=f[j-1]+1;}if (cost_a+before[j-1]>m&&f[j-1]+2<=f[i]){if(f[j-1]+2==f[i])before[i]=min(before[i],cost_b);elsebefore[i]=cost_b;f[i]=f[j-1]+2;}}}printf("%d\n",f[p]);return 0;
}

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Problem Solving（POJ-3265）

Problem Description

Input

Output

Sample Input

Sample Output

Source Program

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