传送门
题目描述
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.
思路
首先发现最开始的一步肯定是贪心，按容量sort一遍可以得到第一个答案，然后稍微转化一下题意，就可以发现，转化后的提议是这个样子：从n个瓶子里面选出num个，使得这num个瓶子的容积之和要大于水的体积，并且当时间最短时，一定有这num个瓶子里的水体积和最大，就可以发现是一个01背包问题。
dp[i][j]表示选了i个瓶子，一共的容积为j的时候，这i个瓶子里水的最大值。由此可得dp方程
dp[j][k]=max(dp[j][k],dp[j-1][k-b[i].v]+b[i].r);
小细节
差点被坑死在这里，按照容量枚举k的时候一定要倒着枚举，不然就成完全背包了2333333。
而且要注意枚举的顺序，最外层枚举的是选第i个，然后枚举容量，然后在枚举选了多少个，这样可以保证一个瓶子只选一次，不然会挂得很惨qwq
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct nod{
int r;
int v;
};
nod b[105];
inline int cmp(nod x,nod y){
return x.v>y.v;
}
int dp[105][20000];
int main(){
int n;
cin>>n;
int rest=0;
for(register int i=1;i<=n;i++){
cin>>b[i].r;
rest+=b[i].r;
}
for(register int i=1;i<=n;i++){
cin>>b[i].v;
}
sort(b+1,b+n+1,cmp);
memset(dp,-1,sizeof(dp)); dp[0][0]=0;
int num=-1,tot=0;
for(register int i=1;i<=n;i++){
tot+=b[i].v;
if(tot>=rest&&num==-1)num=i;
}
cout<<num<<endl;
for(register int i=1;i<=n;i++){ //xuan di ji ge
for(register int k=tot;k>=b[i].v;k--){
for(register int j=1;j<=num;j++){ //xuan le ji ge
if(dp[j-1][k-b[i].v]!=-1){
dp[j][k]=max(dp[j][k],dp[j-1][k-b[i].v]+b[i].r);
//cout<<dp[j-1][k-b[i].v]<<‘ ‘<<b[i].r<<endl;
//cout<<i<<‘ ‘<<j<<‘ ‘<<k<<endl<<endl;
}
}
}
}
// while(1){
// int x,y;
// cin>>x>>y;
// cout<<dp[x][y]<<endl;
// }
int ans=-1;
for(register int i=rest;i<=tot;i++){
ans=max(ans,dp[num][i]);
}
//cout<<tot<<endl;
cout<<rest-ans<<endl;
}
原文地址：https://www.cnblogs.com/Fang-Hao/p/8460595.html