Q - AreYouBusy(混合背包)

野性酷女 2022-02-23 07:24 171阅读 0赞

### 题目描述： ###

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

##### 分析： #####

本题的话，是一个混合背包问题。当s为0的时候，是至少选取一个物品。  
所以状态转移方程为dp\[i\]\[j\]=max\{dp\[i\]\[j\]，dp\[i\]\[j-w\[k\]+p\[k\]，dp\[i-1\]\[j-w\[k\]+p\[k\]\}；  
当s为1的时候，是最多选一个物品。  
所以状态转移方程为  
dp\[i\]\[j\]=max\{dp\[i\]\[j\]，dp\[i-1\]\[j-w\[k\]+p\[k\]\}；  
当s=2的时候，可以看成是一个01背包问题了。  
dp\[i\]\[j\]=max\{dp\[i\]\[j\]，dp\[i-1\]\[j-w\[i\]+p\[k\]，dp\[i\]\[j-w\[k\]\]+p\[k\]\}；

注意当s为0的时候，dp初始值应为-INF；  
当s为1和2时候，dp初始值应该是上一次的dp值。  
//第一篇参考博客里面的评论讲了一个神奇的wa点。  
参考博客：[https://blog.csdn.net/zhou\_yujia/article/details/49637975\#commentsedit][https_blog.csdn.net_zhou_yujia_article_details_49637975_commentsedit]  
[https://blog.csdn.net/u013480600/article/details/40652673][https_blog.csdn.net_u013480600_article_details_40652673]

#include"stdio.h"
    #include"string.h"
    #include"vector"
    #include"functional"
    #include"iostream"
    #include"algorithm"
    using namespace std;
    typedef long long ll;
    #define scanll(a,b) scanf("%I64d%I64d",&a,&b);
    #define scanl(a) scanf("%I64d",&a);
    #define scanff(a,b) scanf("%lf%lf",&a,&b);
    #define scan1f(a) scanf("%lf",&a);
    #define prinll(a,b) printf("%I64d %I64d",a,b);
    #define prinl(a) printf("%I64d",a);
    #define printff(a,b) printf("%lf %lf",a,b);
    #define printlf(a) printf("%lf",a);
    #define OK printf("\n");
    #define MAXSIXE 10010
    #define INF 10001010
    ll N,M;
    ll dp[1100][110];
    int main()
    {
        while(~scanf("%I64d%I64d",&N,&M))
        {
            memset(dp,0,sizeof(dp));
            for(ll i=1; i<=N; i++)
            {
                ll n,id;
                scanll(n,id);
                ll w[110],p[110];
                for(ll i=1; i<=n; i++)
                {
                    scanll(w[i],p[i]);
                }
                if(id==0)
                {
                    for(ll j=0; j<=M; j++)
                        dp[i][j]=-INF;
                    for(ll k=1; k<=n; k++)
                    {
                        for(ll j=M; j>=w[k]; j--)
                        {
                            dp[i][j]=max(dp[i][j],dp[i][j-w[k]]+p[k]);
                            dp[i][j]=max(dp[i][j],dp[i-1][j-w[k]]+p[k]);
                        }
                    }
                }
                else if(id==1)
                {
                    for(ll j=0; j<=M; j++)
                        dp[i][j]=dp[i-1][j];
                    for(ll k = 1; k <= n; k ++)
                    {
                        for(ll j=M; j>=w[k]; j--)
                        {
                            dp[i][j]=max(dp[i][j],dp[i-1][j-w[k]]+p[k]);
                        }
                    }
                }
                else if(id==2)
                {
                    for(ll j=0; j<=M; j++)
                        dp[i][j]=dp[i-1][j];
                    for(ll k=1;k<=n;k++)
                        for(ll j=M;j>=w[k];j--)
                        {
                             dp[i][j]=max(dp[i][j],dp[i][j-w[k]]+p[k]);
                             dp[i][j]=max(dp[i][j],dp[i-1][j-w[k]]+p[k]);
                        }
                }
            }
            printf("%lld\n",max((ll)-1,dp[N][M]));
        }
        return 0;
    }

[https_blog.csdn.net_zhou_yujia_article_details_49637975_commentsedit]: https://blog.csdn.net/zhou_yujia/article/details/49637975#commentsedit
[https_blog.csdn.net_u013480600_article_details_40652673]: https://blog.csdn.net/u013480600/article/details/40652673