*寒假水114——贪心【状压】

最新推荐文章于 2021-02-09 15:43:17 发布

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本文介绍了一个基于状态压缩的动态规划算法，用于解决学生如何合理安排多门课程作业以最小化因延迟提交而被扣除的成绩的问题。该算法通过计算不同作业完成状态下的最小扣分来寻找最优解。

Ignatius has just come back school from the 30th ACM/ICPC. Now he has a lot of homework to do. Every teacher gives him a deadline of handing in the homework. If Ignatius hands in the homework after the deadline, the teacher will reduce his score of the final test. And now we assume that doing everyone homework always takes one day. So Ignatius wants you to help him to arrange the order of doing homework to minimize the reduced score.

InputThe input contains several test cases. The first line of the input is a single integer T that is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=1000) which indicate the number of homework.. Then 2 lines follow. The first line contains N integers that indicate the deadlines of the subjects, and the next line contains N integers that indicate the reduced scores.
OutputFor each test case, you should output the smallest total reduced score, one line per test case.
Sample Input

3
3
3 3 3
10 5 1
3
1 3 1
6 2 3
7
1 4 6 4 2 4 3
3 2 1 7 6 5 4

Sample Output

0
3
5

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
#define max (1<<15)+5

using namespace std;

const int inf=1<<30;
int n,m;
int d[20],f[20];
int pre[max],t[max],dp[max];
char a[20][110];

void output(int x)
{
	if(!x)
		return ;
	output(x-(1<<pre[x]));
	printf("%s\n",a[pre[x]]);
}

int main ()
{
	int T,i,j;
	scanf("%d",&T);
	while(T--)
	{
		cin>>n;
		for(i=0;i<n;i++)
		{
			cin>>a[i]>>d[i]>>f[i];
		}
		m=1<<n;
		for(i=1;i<m;i++)
		{
			dp[i]=inf;
			for(j=n-1;j>=0;j--)
			{
				int temp=1<<j;
				if((temp&i)==0)
					continue;
				int dec=t[i-temp]+f[j]-d[j];
				if(dec<0)
					dec=0;
				if(dp[i]>dp[i-temp]+dec)
				{
					dp[i]=dp[i-temp]+dec;
					t[i]=t[i-temp]+f[j];
					pre[i]=j;
				}
			}

		}
		cout<<dp[m-1]<<endl;
		output(m-1);
	}
	return 0;
}

题解：

假设dp[ _ , _ , _ , …… , _ , _ , _ ]记录了：n门课，在一些完成了，一些未完成的状态下，当前已经花去的时间（time），以及，当前已经被扣掉的最少分数（rs）；

例如：有3门课，那么dp[0,0,0]代表了三门课作业还未做完的状态，dp[0,1,1]代表第二门和第三门的作业已经做完，第一门课作业还没做完的状态；

那么，我们状态转移方程：