C1475D Cleaning the Phone 题解

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Cleaning the Phone

题面翻译

题目大意：

有 $n$ 个物品和一个最低价值 $m$ ，每一个物品有一个价值 $a_i$ 和 体积 $b_i$ 体积只有可能为 $1$ 或 $2$ 。

你需要选出几个物品，使得它们的价值和大于等于 $m$ 且使体积最小。

$T$组询问。

数据范围：

$1\le t\le 10^4$， $1\le n\le 2\cdot 10^5$ ，$1\le a_i\le 10^9$， $1\le b_i\le 2$。

保证 $\sum n\le 2\cdot 10^5$。

题目描述

Polycarp often uses his smartphone. He has already installed $n$ applications on it. Application with number $i$ takes up $a_i$ units of memory.

Polycarp wants to free at least $m$ units of memory (by removing some applications).

Of course, some applications are more important to Polycarp than others. He came up with the following scoring system — he assigned an integer $b_i$ to each application:

• $b_i=1$ — regular application;
• $b_i=2$ — important application.

According to this rating system, his phone has $b_1+b_2+\dots +b_n$ convenience points.

Polycarp believes that if he removes applications with numbers $i_1,i_2,\dots ,i_k$ , then he will free $a_{i_1}+a_{i_2}+\dots +a_{i_k}$ units of memory and lose $b_{i_1}+b_{i_2}+\dots +b_{i_k}$ convenience points.

For example, if $n=5$ , $m=7$ , $a=[5,3,2,1,4]$ , $b=[2,1,1,2,1]$ , then Polycarp can uninstall the following application sets (not all options are listed below):

• applications with numbers $1,4$ and $5$ . In this case, it will free $a_1+a_4+a_5=10$ units of memory and lose $b_1+b_4+b_5=5$ convenience points;
• applications with numbers $1$ and $3$ . In this case, it will free $a_1+a_3=7$ units of memory and lose $b_1+b_3=3$ convenience points.
• applications with numbers $2$ and $5$ . In this case, it will free $a_2+a_5=7$ memory units and lose $b_2+b_5=2$ convenience points.

Help Polycarp, choose a set of applications, such that if removing them will free at least $m$ units of memory and lose the minimum number of convenience points, or indicate that such a set does not exist.

输入格式

The first line contains one integer $t$ ( $1\le t\le 10^4$ ) — the number of test cases. Then $t$ test cases follow.

The first line of each test case contains two integers $n$ and $m$ ( $1\le n\le 2\cdot 10^5$ , $1\le m\le 10^9$ ) — the number of applications on Polycarp’s phone and the number of memory units to be freed.

The second line of each test case contains $n$ integers $a_1,a_2,\dots ,a_n$ ( $1\le a_i\le 10^9$ ) — the number of memory units used by applications.

The third line of each test case contains $n$ integers $b_1,b_2,\dots ,b_n$ ( $1\le b_i\le 2$ ) — the convenience points of each application.

It is guaranteed that the sum of $n$ over all test cases does not exceed $2\cdot 10^5$ .

输出格式

For each test case, output on a separate line:

• -1, if there is no set of applications, removing which will free at least $m$ units of memory;
• the minimum number of convenience points that Polycarp will lose if such a set exists.

样例 #1

样例输入 #1

5
5 7
5 3 2 1 4
2 1 1 2 1
1 3
2
1
5 10
2 3 2 3 2
1 2 1 2 1
4 10
5 1 3 4
1 2 1 2
4 5
3 2 1 2
2 1 2 1

样例输出 #1

2
-1
6
4
3

提示

In the first test case, it is optimal to remove applications with numbers $2$ and $5$ , freeing $7$ units of memory. $b_2+b_5=2$ .

In the second test case, by removing the only application, Polycarp will be able to clear only $2$ of memory units out of the $3$ needed.

In the third test case, it is optimal to remove applications with numbers $1$ , $2$ , $3$ and $4$ , freeing $10$ units of memory. $b_1+b_2+b_3+b_4=6$ .

In the fourth test case, it is optimal to remove applications with numbers $1$ , $3$ and $4$ , freeing $12$ units of memory. $b_1+b_3+b_4=4$ .

In the fifth test case, it is optimal to remove applications with numbers $1$ and $2$ , freeing $5$ units of memory. $b_1+b_2=3$ .

算法：贪心

注意：$1\le m\le 10^9$ 。

动态规划肯定超时了，就往贪心方面去想。

又注意到 $b$ 的取值只能为 $1$ 或$2$，比较特殊，就从中入手。

当总价值 $<m$ 时，肯定不成立，直接输出。

接下来贪心：

因为 $1+1=2$，所以我们可以通过两个体积为 $1$ 的物品与一个体积为 $2$ 的物品（总体积一样）比较价值大小来取。

因此输入时应该根据体积来分别保存到两个数组中，并且排序（从大到小）。之后就可以按照以上步骤来贪心啦~

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const ll N=2e5+10;
ll T,n,m,a[N],b,t1,t2,q1[N],q2[N],ct,ans,h1,h2;
bool cmp(ll l,ll r){return l>r;
}
int main(){ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cin>>T;while(T--){cin>>n>>m;for(int i=1;i<=n;i++) cin>>a[i];t1=t2=ct=ans=0,h1=h2=1;for(int i=1;i<=n;i++){cin>>b;if(b==1) q1[++t1]=a[i];else q2[++t2]=a[i];}sort(q1+1,q1+t1+1,cmp);sort(q2+1,q2+t2+1,cmp);while(ct<m){//贪心取最大值if(!q1[h1]&&!q2[h2]){ans=-1;break;}//q1和q2都取完了,总价值还没达到mif(ct+q1[h1]>=m){ans++;break;}//取q1[h1]后达到if(q1[h1]+q1[h1+1]>=q2[h2]) ct+=q1[h1++],ans++;//取q1else ct+=q2[h2++],ans+=2;//取q2}cout<<ans<<"\n";for(int i=1;i<=n;i++) q1[i]=q2[i]=0;}return 0;
}

感谢大家的支持~