HDU 3535 AreYouBusy

题目链接：​​传送门​​ 题面：

AreYouBusy

Problem Description

Happy New Term!
As having become a junior, xiaoA recognizes that there is not much time for her to AC problems, because there are some other things for her to do, which makes her nearly mad.
What’s more, her boss tells her that for some sets of duties, she must choose at least one job to do, but for some sets of things, she can only choose at most one to do, which is meaningless to the boss. And for others, she can do of her will. We just define the things that she can choose as “jobs”. A job takes time , and gives xiaoA some points of happiness (which means that she is always willing to do the jobs).So can you choose the best sets of them to give her the maximum points of happiness and also to be a good junior(which means that she should follow the boss’s advice)?

Input

There are several test cases, each test case begins with two integers n and T (0<=n,T<=100) , n sets of jobs for you to choose and T minutes for her to do them. Follows are n sets of description, each of which starts with two integers m and s (0<m<=100), there are m jobs in this set , and the set type is s, (0 stands for the sets that should choose at least 1 job to do, 1 for the sets that should choose at most 1 , and 2 for the one you can choose freely).then m pairs of integers ci,gi follows (0<=ci,gi<=100), means the ith job cost ci minutes to finish and gi points of happiness can be gained by finishing it. One job can be done only once.

Output

One line for each test case contains the maximum points of happiness we can choose from all jobs .if she can’t finish what her boss want, just output -1 .

Sample Input

3 3
2 1
2 5
3 8
2 0
1 0
2 1
3 2
4 3
2 1
1 1

3 4
2 1
2 5
3 8
2 0
1 1
2 8
3 2
4 4
2 1
1 1

1 1
1 0
2 1

5 3
2 0
1 0
2 1
2 0
2 2
1 1
2 0
3 2
2 1
2 1
1 5
2 8
3 2
3 8
4 9
5 10

Sample Output

5
13
-1
-1

题目大意：

多组数据。有组工作，每组工作有需要花费的时间和能得到的幸福值。有种类型的工作，每组工作中有个作业，如果这组工作的类型为，那就代表你至少选择一个作业来做，类型为代表最多选择一个作业，类型为代表你可以自由选择作业，输出能得到的最大的幸福值，具体输入格式看代码

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <vector>
#include <iomanip>
#define
#define
#define

using namespace std;
int n, V, m, w[A], v[A], type, f[B][B];

int main() {
    while (cin >> n >> V) { //你可以选择n组工作，有V的时间去做
        memset(f, 0, sizeof f);
        for (int i = 1; i <= n; i++) {
            cin >> m >> type; //第n组工作中有m个作业，类型为type
            for (int j = 1; j <= m; j++) cin >> w[j] >> v[j];
            if (!type) {  //至少选择一个作业来做
                for (int j = 0; j <= V; j++) f[i][j] = - (1 << 29); //初值设小
                for (int j = 1; j <= m; j++)
                  for (int k = V; k >= w[j]; k--)
                    f[i][k] = max(f[i][k], max(f[i][k - w[j]] + v[j], f[i - 1][k - w[j]] + v[j]));//可以由上面的状态转移而来
            }
            else if (type == 1) {  //最多选择一个作业
                for (int j = 0; j <= V; j++) f[i][j] = f[i - 1][j];//继承上一个状态
                for (int j = 1; j <= m; j++)
                  for (int k = V; k >= w[j]; k--)
                    f[i][k] = max(f[i][k], f[i - 1][k - w[j]] + v[j]);
            }
            else { //任意选择作业
                for (int j = 0; j <= V; j++) f[i][j] = f[i - 1][j];//继承状态
                for (int j = 1; j <= m; j++)
                  for (int k = V; k >= w[j]; k--)
                    f[i][k] = max(f[i][k], f[i][k - w[j]] + v[j]);
            }
        }
        cout << max(f[n][V], -1) << endl;
    }
}