hdu3535AreYouBusy (分组背包，WA了很多次)

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 , andthe 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
题目意思：用时T，要求得到最大幸福值。第一行给出两个数n,T，接下来有n个小块，每个小块第一行有两个数m,s，s表示这一小块的类型，为0时表示从中最少要选一个工作，不能不选，为1时表示最多从这一小块中选一个工作，可以不选，为2时表示可选可不选。
#include<stdio.h>#include<iostream>#include<string.h>using namespace std;#define N 120int n0,dp[N][N],T;int max(int a,int b,int c){    if(a<b)a=b;    if(a<c)a=c;    return a;}void init(){    memset(dp,-1,sizeof(dp));    for(int i=0;i<N;i++)        dp[0][i]=0;}void count(int ut,int w){    int j,a,b,c;        for(j=T;j>=ut;j--)        {            a=dp[n0][j];             b=dp[n0][j-ut];             c=dp[n0-1][j-ut];//当类型为2时，一定要注意（相当背包里没有放过类型2）            if(b!=-1)b+=w; if(c!=-1) c+=w;            dp[n0][j]=max(a,b,c);        }}void count_1(int ut,int w){    for(int j=T;j>=ut;j--)    {        if(dp[n0][j]<dp[n0-1][j-ut]+w&&dp[n0-1][j-ut]!=-1)        dp[n0][j]=dp[n0-1][j-ut]+w;    }}int main(){    int n,m,tp,cc,gg,tk;    while(scanf("%d%d",&n,&T)>0)    {        n0=0; init();        while(n--)        {            n0++;            scanf("%d%d",&m,&tp);            if(tp!=0)                for(int i=0;i<=T;i++)                dp[n0][i]=dp[n0-1][i];            while(m--)            {                scanf("%d%d",&cc,&gg);                if(tp==1)                count_1(cc,gg);                else                count(cc,gg);            }        }    printf("%d\n",dp[n0][T]);    }}