zoj 2955 Interesting Dart Game 缩小范围+完全背包

最优得分策略

最新推荐文章于 2019-08-27 16:16:42 发布

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最新推荐文章于 2019-08-27 16:16:42 发布

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分类专栏： ZOJ 动态规划文章标签： dp 缩小范围

本文链接：https://blog.youkuaiyun.com/qq_41410799/article/details/88853709

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本文介绍了一种解决特定游戏问题的方法，即玩家如何通过最少次数获得指定分数。面对庞大的数值范围，文章提出了一种优化策略，利用最大得分减少目标值，并通过动态规划算法（完全背包问题）求解最小次数。

题目：

Recently, Dearboy buys a dart for his dormitory, but neither Dearboy nor his roommate knows how to play it. So they decide to make a new rule in the dormitory, which goes as follows:

Given a number N, the person whose scores accumulate exactly to N by the fewest times wins the game.

Notice once the scores accumulate to more than N, one loses the game.

Now they want to know the fewest times to get the score N.

So the task is :
Given all possible dart scores that a player can get one time and N, you are required to calculate the fewest times to get the exact score N.

Input

Standard input will contain multiple test cases. The first line of the input is a single integer T (1 <= T <= 50) which is the number of test cases. And it will be followed by T consecutive test cases.

Each test case begins with two positive integers M(the number of all possible dart scores that a player can get one time) and N. Then the following M integers are the exact possible scores in the next line.

Notice: M (0 < M < 100), N (1 < N <= 1000000000), every possible score is (0, 100).

Output

For each test case, print out an integer representing the fewest times to get the exact score N.
If the score can't be reached, just print -1 in a line.

Sample Input

Sample Output

2
3
-1

思路：

因为n很大，有1e10,直接dp的话行不通，我们可以通过缩小dp范围来进行优化，当大于1e4时，可以减去尽量多的最大数来缩小到1e4，然后进行dp。

dp的话套用完全背包问题。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
using namespace std;
typedef long long ll;
const int maxn=1e4+5;
const int INF=0x3f3f3f3f;
int t;
int n,m;
int a[maxn];
int dp[maxn<<1];
int Max;
int main()
{
    memset (a,0,sizeof(a));
    scanf("%d",&t);
    while(t--)
    {
        Max=-0x3f3f3f3f;
        int ans=0;
        scanf("%d%d",&m,&n);
        for (int i=1;i<=m;i++)
        {
            scanf("%d",&a[i]);
            Max=max(Max,a[i]);
        }
        if(n>10000)
        {
            int t=n-10000;
            ans+=t/Max;
            n=10000+t%Max;
        }
        for (int i=1;i<=n;i++) dp[i]=INF;
        dp[0]=0;
        for (int i=1;i<=m;i++)
        {
            for (int j=a[i];j<=n;j++)
            {
                dp[j]=min(dp[j-a[i]]+1,dp[j]);
            }
        }
        if(dp[n]==INF) printf("-1\n");
        else printf("%d\n",dp[n]+ans);
    }
    return 0;
}