Bottles CodeForces - 730J
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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.
Example
4
3 3 4 3
4 7 6 5
2 6
2
1 1
100 100
1 1
5
10 30 5 6 24
10 41 7 8 24
3 11
Note
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.
题解:dp[ i ][ j ][ k ]表示1~i个瓶子中已选j个瓶子得到的最大实际装水量。那么dp[ i ][ j ][ k ]=max(dp[ i ] [ j ] [ k ],dp[ i - 1 ][ j - 1][ k - b[ i ] ] + a[ i ] )。
首先,题目要求最少的瓶子最短的时间,而最少的瓶子满足其总容量大于->实际水的总量<-(sumN)。所需要的时间,则是水的总量(sumN)减去这k个瓶子装的水的总量。所以time=sum1-maxSum(K)。
dp好难,初始化都不好掌握 ,><
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn=105; 5 6 int n,sum1,sum2; 7 int dp[maxn][maxn*maxn]; 8 9 struct node{ 10 int a,b; 11 bool operator<(const node& i)const{ 12 return i.b<b; 13 } 14 }p[maxn]; 15 16 void Inite(){ 17 memset(dp,-1,sizeof(dp)); 18 dp[0][0]=0; 19 } 20 21 void Solve(){ 22 int tem=0,q=1; 23 for(int i=1;i<=n;i++){ 24 tem+=p[i].b; 25 q=i; 26 if(tem>=sum1) break; 27 } 28 for(int i=1;i<=n;i++) 29 for(int j=q;j>=1;j--) 30 for(int k=sum2;k>=p[i].b;k--) 31 if(~dp[j-1][k-p[i].b]) dp[j][k]=max(dp[j][k],dp[j-1][k-p[i].b]+p[i].a); 32 int ans=0; 33 for(int i=sum1;i<=sum2;i++) ans=max(ans,dp[q][i]); 34 cout<<q<<" "<<sum1-ans<<endl; 35 } 36 37 int main() 38 { 39 cin>>n; 40 Inite(); 41 sum1=sum2=0; 42 for(int i=1;i<=n;i++){ cin>>p[i].a; sum1+=p[i].a; } 43 for(int i=1;i<=n;i++){ cin>>p[i].b; sum2+=p[i].b; } 44 sort(p+1,p+n+1); 45 Solve(); 46 }
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