Doing Homework
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12350 Accepted Submission(s): 5951
Problem Description
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.
Input
The 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.
Output
For 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.
Author
Ignatius.L
Recommend
算法分析：
题意：
有n门课，每门课有截止时间和完成所需的时间，如果超过规定时间完成，每超过一天就会扣1分，问怎样安排做作业的顺序才能使得所扣的分最小
分析：
这道题有一个明显的标志,N为15，一般N为15 16都是用状态压缩DP。
n门课，(1<<n)-1,有n个1，表示n门课都完成。若该位为0，表示这门课没有完成。i从1到1<<n -1，模拟所有情况。dp[i]记录到达状态i扣的最少分，对每种情况dp[i]，假设有m门课完成，肯定是由m-1门状态而来，所以只要遍每门课没完成的的状态，找到最小值。然后+当前课程所扣的分数。
具体实现细节看代码，注释很清楚。
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <queue>
#include <algorithm>
#include <map>
#include <cmath>
#include <iomanip>
#define INF 99999999
typedef long long LL;
using namespace std;
const int MAX=(1<<15)+10;
int n;
int dp[MAX],t[MAX],pre[MAX],dea[20],fin[20];//dp[i]记录到达状态i扣的最少分,t时相应的花去多少天了
char s[20][110];
void output(int x){
if(!x)return;
output(x-(1<<pre[x]));
printf("%s\n",s[pre[x]]);
}
int main(){
int T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i=0;i<n;++i)
scanf("%s%d%d",&s[i],&dea[i],&fin[i]);
int bit=1<<n;
for(int i=1;i<bit;++i){//枚举到达状态i
dp[i]=INF;//初始化到达状态i的扣分
for(int j=n-1;j>=0;--j){//由于输入时按字符大小输入，而每次完成j相当于把j放在后面完成且下面判断是dp[i]>dp[i-temp]+score
int temp=1<<j; //所以是n-1开始，如果下面判断是dp[i]>=dp[i-temp]+score则从0开始
if(!(i&temp))continue;//状态i不存在作业j完成则不能通过完成作业j到达状态i
int score=t[i-temp]+fin[j]-dea[j];//i-temp表示没有完成j的那个状态
if(score<0)score=0;//证明截止时间前就完成了
if(dp[i]>dp[i-temp]+score){//更新最小值
dp[i]=dp[i-temp]+score;
t[i]=t[i-temp]+fin[j];//到达状态i花费的时间
pre[i]=j;//到达状态i的前驱,为了最后输出完成作业的顺序
}
}
}
printf("%d\n",dp[bit-1]);
output(bit-1);//输出完成作业的顺序
}
return 0;
}
