（sgu）Balloons（记忆化搜索）

Time Limit : 1000/500ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Problem Description

Due to the preparation to the wedding you have to fill up M balloons. There are N volunteers and one device for balloons filling up. Device can fill at most one balloon at a time. Each volunteer can continuously operate the device for 1 minute.
Volunteer number i can fill Ai balloons during one minute and then he or she must have at least Bi minutes of rest.
You are to write a program that computes minimal possible time of filling up M balloons.

Input
There are two integers M and N on the first line (1 <= M <= 100, 1 <= N <= 10).
Next N lines contain two integers each: Ai and Bi (0 <= Ai <= 10, 1 <= Bi <= 4).

Output
Print one integer T - the minimal number of minutes required to filling M balloons.

Sample test(s)

Input
10 3
4 4
3 4
2 3

Output
5

题意：N(1<=N<=10)我的一台装置填充M（1<=M<=100）个气球，每一个人一分钟能充Ai个（0<=Ai<=10）,休息Bi（1<=Bi<=4）分钟后才能够继续充气球，文填充M个气球最少须要多少时间？
分析：看到题目，第一想法就是排序，先大后小，而后DFS，这样显然错的，样例都过不了，而后又想着暴力写出来，不行，确定会超时，只给了0.5s。
而后无奈看了大佬写的博客，他就直接观察到了Bi的范围，是1到4，这个很小，能够开一个五维数组ans[sum][a][b][c][d]来表示还剩sum个气球，前一分钟是a在充气球，前两分钟是b在充气球，前3分钟是c在充气球，前4分钟是d在吹气球，而后直接记忆化搜索

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
const int N=11,INF=0x3f3f3f3f;
int ans[101][N][N][N][N],A[N],B[N];
int n,m,res;
int dfs(int sum,int a,int b,int c,int d)
{
    if(sum<=0) sum=0;///所有充完
    if(ans[sum][a][b][c][d]!=INF) return ans[sum][a][b][c][d];///还剩sum个球所用的最短期已找到
    if(sum==0) ans[sum][a][b][c][d]=0;///所有填充完毕
    else
    {
        int flag=0;
        for(int i=1; i<=n; i++)
        {
            if(a==i&&B[i]>0) continue;
            if(b==i&&B[i]>1) continue;
            if(c==i&&B[i]>2) continue;
            if(d==i&&B[i]>3) continue;
            flag=1;
            int t=dfs(sum-A[i],i,a,b,c);///第i人在充气球
            ans[sum][a][b][c][d]=min(ans[sum][a][b][c][d],t+1);///+1是由于最后还有一我的在充
        }
        if(!flag)///都在休息
        {
            int t=dfs(sum,0,a,b,c);
            ans[sum][a][b][c][d]=min(ans[sum][a][b][c][d],t+1);
        }
    };
    return ans[sum][a][b][c][d];
}
int main()
{
    scanf("%d%d",&m,&n);
    memset(ans,INF,sizeof(ans));
    for(int i=1; i<=n; i++)
        scanf("%d%d",&A[i],&B[i]);
    res=dfs(m,0,0,0,0);
    printf("%d\n",res);
    return 0;
}

写这题的时候那个oj的判题有问题，我写错了一个地方，致使样例都过不了，还AC了。。。。。 不过这个思路应该没问题
若有不对的地方，烦请大佬指出，谢谢
小结：像这类题，观察数据范围，好比数组B，范围小，就能开一个数组记忆最少时间，而后和DFS结合。
本身对记忆化搜索不太感冒，还要勤加练习。