HDU 3535 AreYouBusy

点击打开链接

 

AreYouBusy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2057    Accepted Submission(s): 773

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 32 12 53 82 01 02 13 24 32 11 13 42 12 53 82 01 12 83 24 42 11 11 11 02 15 32 01 02 12 02 21 12 03 22 12 11 52 83 23 84 95 10

 

 

Sample Output

513-1-1

 

 

Author

hphp

 

 

Source

2010 ACM-ICPC Multi-University Training Contest（10）——Host by HEU

 

题意：有n组工作，T时间，每个工作组中有m个工作，改组分类是s，s是0是组内至少要做一件，是1时最多做一件，2时随意，每项工作的描述是花费的时间和获得的快乐值，求在T时间内可获的最大快乐值。

思路：当s=0时，定义dp数组除了0行以外都定义成-1，这样就保证了必须时被选中的物品不会被后面组中的物品覆盖。

s=1时，最多选一项要么不选。所以状态转移方程：dp[i][k]=max{ dp[i][k],dp[i-1][k-cost[j]]+val[k]}。由于要保证全局的最优解，所以要将上一组的dp值覆盖到这一组上。

s=2时，任意选，选不选无所谓，状态转移方程:dp[i][k]=max{ dp[i][k],dp[i-1][k-cost[j]]+val[k],dp[i][k-cost[j]]+val[j] }。为了得到全局的最优解也要将上一组的值覆盖到这一组上。

 

#include<stdio.h>#include<string.h>int c[1007],w[1007],dp[1007][1007];int max(int a,int b){    return a>b?a:b;}int main(){    int n,t,i,j,k;    while(scanf("%d%d",&n,&t)!=EOF)    {        memset(dp,-1,sizeof(dp));        memset(dp[0],0,sizeof(dp[0]));        for(i=1;i<=n;i++)        {            int m,s;            scanf("%d%d",&m,&s);            for(j=0;j<m;j++)            {                scanf("%d%d",&c[j],&w[j]);            }            if(s==0)            {                for(j=0;j<m;j++)                for(k=t;k>=c[j];k--)                {                    if(dp[i][k-c[j]]!=-1)                    {                        dp[i][k]=max(dp[i][k],dp[i][k-c[j]]+w[j]);                    }                    if(dp[i-1][k-c[j]]!=-1)                    {                        dp[i][k]=max(dp[i][k],dp[i-1][k-c[j]]+w[j]);                    }                }            }            else if(s==1)            {                for(j=0;j<=t;j++)                dp[i][j]=dp[i-1][j];                for(k=0;k<m;k++)                {                    for(j=t;j>=c[k];j--)                    {                        if(dp[i-1][j-c[k]]!=-1)                        dp[i][j]=max(dp[i][j],dp[i-1][j-c[k]]+w[k]);                    }                }            }            else            {                for(j=0;j<=t;j++)                dp[i][j]=dp[i-1][j];                for(k=0;k<m;k++)                for(j=t;j>=c[k];j--)                {                    if(dp[i][j-c[k]]!=-1)                    dp[i][j]=max(dp[i][j],dp[i][j-c[k]]+w[k]);                }            }        }        printf("%d\n",dp[n][t]);    }    return 0;}