ACM Button Bashing

题目：

You recently acquired a new microwave, and noticed that it provides a large number of buttons to be able to quickly specify the time that the microwave should be running for. There are buttons both for adding time, and for subtracting time. You wonder how efficient you can be when entering cooking times: you want to minimize the number of required button presses.

The microwave can be running for at least $0$seconds, and at most $1$ hour. If a button press would result in a cooking time of less than $0$seconds, the microwave will set the cooking time to $0$ seconds. If a button press would result in a cooking time of more than $1$ hour, the microwave will set the cooking time to $1$hour. Initially, the microwave will run for $0$seconds. There will always be a button adding at least $1$ second to the cooking time.

Given the buttons that the microwave provides for entering cooking times, determine the least amount of button presses required to let the microwave run for a certain amount of time. If it is not possible to enter the desired cooking time precisely, determine the smallest achievable cooking time above the target, and the minimum number of button presses required for that cooking time, instead. The microwave does not allow to adjust the cooking time once it has started cooking.

Input Format

On the first line one positive number: the number of test cases, at most $100$. After that per test case:

• one line with two space-separated integers $n$ and $t$ $(1\le n\le 16,0\le t\le 3600)$: the number of buttons available to change the cooking time, and the desired cooking time in seconds, respectively.
• one line with $n$ space-separated integers b_ibi​ (-3600 \le b_i \le 3600)(−3600≤bi​≤3600): the number of seconds added to the cooking time when button $i$ is pressed.

Output Format

Per test case:

• one line with two space-separated integers: the minimum number of button presses required to reach the required cooking time, and the minimum number of extra seconds that the microwave must be running for, respectively.

样例输入

2
3 50
-10 10 60
1 50
20

样例输出

2 0
3 10

题意：

给出微波炉需要的定时，以及n个不同的按钮，不同的按钮对应不同的效果，增加或者减少一些时间，当然微波炉最多运行一个小时；

分析：首先想到用dp做，因为最近一直在学dp，但是发现有负数，不好判断，还是bfs吧；

要注意的点就是当时间<0或者>3600时，一定要写成=0或者=3600；

第二点，book数组定义时一定要定义为一个非常大的数，分析见代码；

代码：

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
int main()
{
    int i,j,n,m,k,num[20],book[4000];
    scanf("%d",&k);
    while(k--)
    {
        queue<int>q;
        memset(book,INF,sizeof(book));
        scanf("%d %d",&n,&m);
        for(i=1;i<=n;i++)
            scanf("%d",&num[i]);
        q.push(0);
        book[0]=0;
        while(!q.empty())
        {
            int head=q.front();
            q.pop();
            for(i=1;i<=n;i++)
            {
               int tail=head+num[i];
                if(tail<0)
                    tail=0;
                else if(tail>3600)
                    tail=3600;
                if(book[tail]<=book[head]+1)由于判断条件是<=，因此一定要初始化为一个非常大的数
                    continue;
                q.push(tail);
                book[tail]=book[head]+1;
            }
        }
        int t;
        for(t=m;t<=3600;t++)//这个是从以为大佬那里学来的
        {
            if(book[t]!=INF)
                break;
        }
        printf("%d %d\n",book[t],t-m);
    }
    return 0;
}