HDU 1074 Doing Homework(状态压缩 + DP）

标签：acm   dp   压缩   

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

#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#define LL long long
using namespace std;
const int MAXN = 16;
const int inf = 0x3f3f3f3f;
struct Node
{
    char name[20];
    int D, C;
}node[MAXN];
int dp[1<<MAXN];
int pre[1<<MAXN];
int n;
void output(int status)//按字典序输出(题目输入顺序即是字典序)
{
    if(status == 0) return ;
    int t = 0;
    for(int i=0;i<n;i++)
    {
        if((status & (1<<i)) != 0 && (pre[status] & (1<<i)) == 0)
        {
            t = i;
            break;
        }
    }
    output(pre[status]);
    printf("%s\n", node[t].name);
}
int main()
{
    int T;
    scanf("%d", &T);
    while(T--)
    {
        scanf("%d", &n);
        for(int i=0;i<n;i++)
        {
            scanf("%s%d%d", &node[i].name, &node[i].D, &node[i].C);
        }
        for(int i=0;i<(1<<n);i++) dp[i] = inf;
        dp[0] = 0;//初始化
        for(int i=0;i<(1<<n);i++)
        {
            for(int j=0;j<n;j++)
            {
                if(i & (1<<j)) continue;//如果该作业已经完成则继续
                int time = 0;
                for(int k=0;k<n;k++) if(i & (1<<k))
                    time += node[k].C;
                time += node[j].C;//统计直到完成该作业所需的总时间
                if(time > node[j].D) time -= node[j].D;
                else time = 0;
                if(dp[i|(1<<j)] > dp[i] + time)
                   {
                       dp[i|(1<<j)] = dp[i] + time;
                       pre[i|(1<<j)] = i;
                   }
            }
        }
        printf("%d\n", dp[(1<<n)-1]);
        output((1<<n)-1);
    }
    return 0;
}

HDU 1074 Doing Homework(状态压缩 + DP）

标签：acm   dp   压缩   

原文地址：http://blog.csdn.net/moguxiaozhe/article/details/44258665