hdoj-1074：
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
题意就是：有n个作业每个作业有一个最后完成时间和一个需要的时间，如果超过最后完成的时间，超过一天扣一分，求最少要扣多少分
思路：根据题意 n最多有15，所以把状态压缩成2进制，每一位0表示还没有选 1 表示选了，状态转移就是 前一个转态到后一个 根据二进制的性质
，最后还要输出路径，所以要 保留他，用一个结构体保留他的前驱和他由前驱到现在 需要完成那个作业;
代码；
1 #include <stdlib.h> 2 #include<iostream> 3 #include<algorithm> 4 #include<bits/stdc++.h> 5 #include<cstdio> 6 #include<cstring> 7 #include<cmath> 8 #define ll long long 9 const ll INF=0x3f3f3f3f; 10 #define mod 1000000007 11 #define mem(a,b) memset(a,b,sizeof(a)) 12 #define INF 0x3f3f3f3f 13 //__builtin_popcount 14 using namespace std; 15 //priority_queue 16 const ll MAX=100000; 17 struct A 18 { 19 string name; 20 int a,b; 21 } s[30]; 22 struct B 23 { 24 int time,pre,cot,id;// 25 } dp[1<<15+3]; 26 void prin(int z) 27 { 28 if(z) 29 { 30 prin(dp[z].pre); 31 cout<<s[dp[z].id].name<<endl; 32 } 33 } 34 int main() 35 { 36 int t; 37 cin>>t; 38 while(t--) 39 { 40 mem(dp,0); 41 int n; 42 int m; 43 cin>>n; 44 m=1<<n; 45 for(int i=1; i<=n; i++) 46 { 47 cin>>s[i].name>>s[i].a>>s[i].b; 48 } 49 for(int i=1; i<m; i++) 50 { 51 dp[i].cot=INF; 52 for(int j=0; j<n; j++) 53 { 54 if(((1<<j)&i)==0||(1<<j)>i)continue; 55 int pre=i-(1<<j); 56 int tt=dp[pre].time+s[j+1].b-s[j+1].a; 57 tt=max(tt,0); 58 if(dp[pre].cot+tt<=dp[i].cot) 59 { 60 dp[i].cot=dp[pre].cot+tt; 61 dp[i].id=j+1; 62 dp[i].time=dp[pre].time+s[j+1].b; 63 dp[i].pre=pre; 64 } 65 } 66 } 67 cout<<dp[m-1].cot<<endl; 68 prin(m-1); 69 } 70 }