hdu 3535 混合背包

AreYouBusy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4087    Accepted Submission(s): 1627

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

 

题意：

老板让小A做一些工作，有多个case

每个case以 n T 开始，n表示有多少类工作，T表示限制的时间

每一类工作以m  S开头，m表示这一类里面有m种工作

s==0 表示至少要选择一种工作做

s==1 表示最多要选择一种工作做

s==2 表示可以任意选择工作去做

每一种工作有两个数字c  g，c表示需要花多少时间，g表示可以得到多少点快乐数

求解在规定时间内完成老师给的任务，最大的快乐数

如若不能完成就输出-1

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int dp[110][110];int main(){    int n,t;    //freopen("in.txt","r",stdin);    while(scanf("%d%d",&n,&t)!=EOF)    {        int m,s;        memset(dp,0,sizeof(dp));        for(int i=1;i<=n;i++){            scanf("%d%d",&m,&s);            if(!s)                for(int k=0;k<=t;k++)                    dp[i][k]=-1;            else                for(int k=0;k<=t;k++)                    dp[i][k]=dp[i-1][k];            int c,g;            for(int j=1;j<=m;j++){                scanf("%d%d",&c,&g);                for(int k=t;k>=c;k--){                    if(s==0){                        if(dp[i][k-c]!=-1)                            dp[i][k]=max(dp[i][k],dp[i][k-c]+g);                        if(dp[i-1][k-c]!=-1)                            dp[i][k]=max(dp[i][k],dp[i-1][k-c]+g);                    }                    else if(s==1){                        if(dp[i-1][k]!=-1)                            dp[i][k]=max(dp[i][k],dp[i-1][k]);                        if(dp[i-1][k-c]!=-1)                            dp[i][k]=max(dp[i][k],dp[i-1][k-c]+g);                    }                    else{                        if(dp[i-1][k]!=-1)                            dp[i][k]=max(dp[i][k],dp[i-1][k]);                        if(dp[i][k-c]!=-1)                            dp[i][k]=max(dp[i][k],dp[i][k-c]+g);                        if(dp[i-1][k-c]!=-1)                            dp[i][k]=max(dp[i][k],dp[i-1][k-c]+g);                    }                }            }        }        printf("%d\n",dp[n][t]);    }    return 0;}