Doing Homework （状态DP&&位运算）

最新推荐文章于 2021-06-04 10:14:26 发布

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#状态DP

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神奇的位运算/二进制

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本篇介绍了一个算法问题：如何为刚参加完ACM/ICPC比赛的学生Ignatius合理安排大量课后作业的完成顺序，以最小化因延期提交而被扣分的情况。文章提供了一种解决方案，通过动态规划来确定最佳作业完成顺序。

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, 1 day for 1 point. And as you know, doing homework always takes a long time. 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 which is the number of test cases. T test cases follow.
Each test case start with a positive integer N(1<=N<=15) which indicate the number of homework. Then N lines follow. Each line contains a string S(the subject's name, each string will at most has 100 characters) and two integers D(the deadline of the subject), C(how many days will it take Ignatius to finish this subject's homework).

Note: All the subject names are given in the alphabet increasing order. So you may process the problem much easier.
OutputFor each test case, you should output the smallest total reduced score, then give out the order of the subjects, one subject in a line. If there are more than one orders, you should output the alphabet smallest one.
Sample Input

2
3
Computer 3 3
English 20 1
Math 3 2
3
Computer 3 3
English 6 3
Math 6 3

Sample Output

2
Computer
Math
English
3
Computer
English
Math

Hint

In the second test case, both Computer->English->Math and Computer->Math->English leads to reduce 3 points, but the 
word "English" appears earlier than the word "Math", so we choose the first order. That is so-called alphabet order.

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<stack>
typedef long long LL;
using namespace std;
const LL inf=1000000000;
int t;
int n;
struct node1
{
    char name[105];
    int d,c;
}sub[20];

struct node2
{
    LL time,cost,pre,now;
}dp[1<<16];

int main()
{
    scanf("%d",&t);
    while(t--)
    {
        memset(dp,0,sizeof(dp));
        dp[1].pre=0;
        scanf("%d",&n);
        for(int i=0;i<n;i++){
            scanf("%s%d%d",&sub[i].name,&sub[i].d,&sub[i].c);
        }
        for(int i=1;i<(1<<n);i++){
            dp[i].cost=inf;
            for(int j=n-1;j>=0;j--){
                if(i&(1<<j)){//是否存在该状态
                    LL ii=(i&(~(1<<j)));//当这门课完成之前的状态(也就是把i状态（2进制）的第j个位置上的数变为0）
                    LL s=dp[ii].time+sub[j].c-sub[j].d;
                    if(s<0)
                        s=0;
                    if(s+dp[ii].cost<dp[i].cost){
                        dp[i].time=dp[ii].time+sub[j].c;
                        dp[i].pre=ii;
                        dp[i].now=j;
                        dp[i].cost=s+dp[ii].cost;
                    }
                }
            }
        }
        stack<LL>st;
        printf("%lld\n",dp[(1<<n)-1].cost);
        LL a=((1<<n)-1);//最终状态是有n个1的状态
        while(a){
            st.push(dp[a].now);
            a=dp[a].pre;
        }
        while(!st.empty()){
            printf("%s\n",sub[st.top()].name);
            st.pop();
        }
    }
}