HDU 1074 Doing Homework(经典状压dp)

 

题目链接

 Doing Homework       

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：有N门功课，分别有deadline（最后期限）和cost（做这门作业要花的时间），在ddl后x天交作业就要扣x分，而且不能同时做两门功课及以上，问怎么安排顺序扣的分最少。

全排列的话就是N!，肯定会TLE。

相对而言N其实很小，那么可以枚举每门功课选不选，那么状态数最多只有2^15个。这样就不会tle了。

状态压缩DP就是一种用二进制来替代枚举的过程(虽说我感觉本质还是暴力枚举

常使用位运算

dp[i|(1<<j)] = max(dp[i|(1<<j)] , dp[i] + sum_cost) 其中sum_cost表示做完这些已选中的科目所需的时间

输出还有一个打印顺序的，就用pre数组来记录就好了。

 

代码如下：

 

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int INF = 0x3f3f3f3f;
const int maxn = 16;
struct Node {
    char name[110];
    int D,C;
} node[maxn];
int dp[1<<maxn], pre[1<<maxn];
int n;

void print_ans(int now) {
    if (!now) return;
    int temp;
    for (int i = 0; i < n; i++) {
        if ((now & (1<<i)) && !(pre[now] & (1<<i))) {
            temp = i;
            break;
        }
    }
    print_ans(pre[now]);
    puts(node[temp].name);
}

int main() {
    int T; scanf("%d", &T);
    while (T--) {
        memset(dp, 0x3f, sizeof(dp));
        memset(pre, 0, sizeof(pre));
        scanf("%d", &n);
        for (int i = 0; i < n; i++) scanf("%s%d%d", node[i].name, &node[i].D, &node[i].C);
        dp[0] = 0;
        for (int i = 0; i < (1 << n); i++) {
            for (int j = 0; j < n; j++) {
                if (i & (1 << j)) continue;
                int s = 0;
                for (int k = 0; k < n; k++)
                    if (i & (1 << k)) s += node[k].C;
                s += node[j].C;
                if (s > node[j].D) s -= node[j].D;
                else s = 0;
                if (dp[i|(1<<j)] > dp[i] + s) {
                    dp[i|(1<<j)] = dp[i] + s;
                    pre[i|(1<<j)] = i;
                }
            }
        }
        printf("%d\n", dp[(1<<n)-1]);
        print_ans((1<<n)-1);
    }
}

View Code

 

转载于:https://www.cnblogs.com/Mrzdtz220/p/10392800.html