状压DP：用到状压的DP。
什么是状压？
把一系列状态压缩成一个状态（通常用二进制）。
例题：HDU1074
Doing Homework
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12604 Accepted Submission(s): 6079
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 3Sample Output
2
Computer
Math
English
3
Computer
English
MathHint
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.
Source
HDU1074
题意：有n个任务，每个任务有一个截止时间，超过截止时间一天，要扣一个分。求如何安排任务，使得扣的分数最少。
分析：可以将每一个任务完成看成1，未完成看成0，则每一种状态可以转换成一个n位二进制，换算成十进制就是0-2^n-1
这样就可以进行dp了（代码中进行了注释）。
代码：
#include <iostream>
#include <cstdio>
const int inf=0x3f3f3f3f;
const int MAX=(1<<15)+10;
using namespace std;//dp记录到达状态i扣的最少的分
//t为达到i花的时间
//pre记录路径
//ddl为截止时间，len为时长
int n,dp[MAX],t[MAX],pre[MAX],ddl[20],len[20];
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],&ddl[i],&len[i]);//设置上限，1<<n表示2^nint bit=1<<n;for (int i=1;i<bit;i++){dp[i]=inf;for (int j=n-1;j>=0;j--){int temp=1<<j;//如果第j位不为1说明不能从i-temp转移过来if (!(i&temp))continue;//扣分为上一个状态的时间+长度-截止时间int score=t[i-temp]+len[j]-ddl[j];//如果发现扣分为负则不需要扣分if (score<0)score=0;if (dp[i]>dp[i-temp]+score){//更新dpdp[i]=dp[i-temp]+score;//更新花的时间t[i]=t[i-temp]+len[j];//记录路径pre[i]=j;}}}printf("%d\n",dp[bit-1]);output(bit-1);}return 0;
}