CodeForces 730J Bottles (DP 01背包)

Bottles
Time Limit:2000MS Memory Limit:524288KB 64bit IO Format:%I64d & %I64u
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Status

Practice

CodeForces 730J
Description
Nick has n bottles of soda left after his birthday. Each bottle is described by two values: remaining amount of soda ai and bottle volume bi (ai ≤ bi).

Nick has decided to pour all remaining soda into minimal number of bottles, moreover he has to do it as soon as possible. Nick spends x seconds to pour x units of soda from one bottle to another.

Nick asks you to help him to determine k — the minimal number of bottles to store all remaining soda and t — the minimal time to pour soda into k bottles. A bottle can’t store more soda than its volume. All remaining soda should be saved.

Input
The first line contains positive integer n (1 ≤ n ≤ 100) — the number of bottles.

The second line contains n positive integers a1, a2, …, an (1 ≤ ai ≤ 100), where ai is the amount of soda remaining in the i-th bottle.

The third line contains n positive integers b1, b2, …, bn (1 ≤ bi ≤ 100), where bi is the volume of the i-th bottle.

It is guaranteed that ai ≤ bi for any i.

Output
The only line should contain two integers k and t, where k is the minimal number of bottles that can store all the soda and t is the minimal time to pour the soda into k bottles.

Sample Input
Input
4
3 3 4 3
4 7 6 5
Output
2 6
Input
2
1 1
100 100
Output
1 1
Input
5
10 30 5 6 24
10 41 7 8 24
Output
3 11
Hint
In the first example Nick can pour soda from the first bottle to the second bottle. It will take 3 seconds. After it the second bottle will contain 3 + 3 = 6 units of soda. Then he can pour soda from the fourth bottle to the second bottle and to the third bottle: one unit to the second and two units to the third. It will take 1 + 2 = 3 seconds. So, all the soda will be in two bottles and he will spend 3 + 3 = 6 seconds to do it.

#include "cstring"
#include "cstdio"
#include "string.h"
#include "iostream"
#include "cmath"
#include "algorithm"
using namespace std;
#define MAX 105
typedef struct node
{
    int left,volume;
}node;
node a[105];
bool cmp(node a,node b)
{
    if(a.volume!=b.volume)
        return a.volume>b.volume;
    else
        return a.left>b.left;
}

int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        //dp[i][j]  选i个瓶子其中自带j的水，最终可以达到的最大水量(选中的的瓶子的总体积)
        int dp[105][10005];
        int sum=0;
        memset(dp,-1,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i].left);
            sum+=a[i].left;
        }
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i].volume);
        int cnt=0;
        int total=0;
        sort(a+1,a+1+n,cmp);
        for(int i=1;i<=n;i++)
        {
            total+=a[i].volume;
            if(total>=sum)
            {
                cnt=i;
                break;
            }
        }
        dp[0][0]=0;
        for(int i=1;i<=n;i++)
        {
            for(int j=sum;(j-a[i].left)>=0;j--)
            {
                for(int k=i;k>=1;k--)
                {
                    if(dp[k-1][j-a[i].left]!=-1)
                        dp[k][j]=max(dp[k][j],dp[k-1][j-a[i].left]+a[i].volume);
                }
            }
        }
        int ans=0;
        for(int i=sum;i>=1;i--)
        {
            if(dp[cnt][i]>=sum)
            {
                ans=sum-i;
                break;
            }
        }
        printf("%d %d\n",cnt,ans);

    }
}