UVA Stamps and Envelope Size

Philatelists have collected stamps since long before postal workers were disgruntled. An excess of stamps may be bad news to a country's postal service, but good news to those that collect the excess stamps. The postal service works to minimize the number of stamps needed to provide seamless postage coverage. To this end you have been asked to write a program to assist the postal service.

Envelope size restricts the number of stamps that can be used on one envelope. For example, if 1 cent and 3 cent stamps are available and an envelope can accommodate 5 stamps, all postage from 1 to 13 cents can be ``covered":

Although five 3 cent stamps yields an envelope with 15 cents postage, it is not possible to cover an envelope with 14 cents of stamps using at most five 1 and 3 cent stamps. Since the postal service wants maximal coverage without gaps, the maximal coverage is 13 cents.

Input

The first line of each data set contains the integer S, representing the maximum of stamps that an envelope can accommodate. The second line contains the integer N, representing the number of sets of stamp denominations in the data set. Each of the next N lines contains a set of stamp denominations. The first integer on each line is the number of denominations in the set, followed by a list of stamp denominations, in order from smallest to largest, with each denomination separated from the others by one or more spaces. There will be at most S denominations on each of the N lines. The maximum value of S is 10, the largest stamp denomination is 100, the maximum value of N is 10.

The input is terminated by a data set beginning with zero (S is zero).

Output

Output one line for each data set giving the maximal no-gap coverage followed by the stamp denominations that yield that coverage in the following format:

max coverage = <value> : <denominations>

If more than one set of denominations in a set yields the same maximal no-gap coverage, the set with the fewest number of denominations should be printed (this saves on stamp printing costs). If two sets with the same number of denominations yield the same maximal no-gap coverage, then the set with the lower maximum stamp denomination should be printed. For example, if five stamps fit on an envelope, then stamp sets of 1, 4, 12, 21 and 1, 5, 12, 28 both yield maximal no-gap coverage of 71 cents. The first set would be printed because both sets have the same number of denominations but the first set's largest denomination (21) is lower than that of the second set (28). If multiple sets in a sequence yield the same maximal no-gap coverage, have the same number of denominations, and have equal largest denominations, then print the set with the lewer second-maximum stamp denomination, and so on.

Sample Input

5
2
4 1 4 12 21
4 1 5 12 28
10
2
5 1 7 16 31 88
5 1 15 52 67 99
6
2
3 1 5 8
4 1 5 7 8
0

Sample Output

max coverage =  71 :  1  4 12 21
max coverage = 409 :  1  7 16 31 88
max coverage =  48 :  1  5  7  8

醉了的一题，卡格式。。。

PE8次。。。

求n个邮票集中连续区间最大的邮票集合

开始想用二维完全背包试试看，状态没有想的太清楚。。。

后来看到别人博客才发现用近似暴力的遍历标记实现起来很清晰，而且据说这是WF的某题=。=

#include <iostream>
#include <stdio.h>
#include <string>
#include <cstring>
#include <algorithm>

using namespace std;
int a[1009];
int dp[1009];
int k;

int b[1009];
int cmax;
int ans;
int minnum;
int minmax;

int main()
{
    int n,m;
    while(scanf("%d",&m),m)
    {
        scanf("%d",&n);

        ans=-1;
         minnum=100000;
         minmax=100000;

     while(n--)
        {
            scanf("%d",&k);
            for(int j=0;j<k;j++)
                scanf("%d",&a[j]);

           memset(dp,0,sizeof dp);
           dp[0]=1;
           int end=0;

           for(int i=1;i<=m;i++)
           {
               for(int j=end;j>=0;j--)
               {
                   if(dp[j])
                   {
                       for(int t=0;t<k;t++)
                       dp[j+a[t]]=1;
                   }
               }
               end+=a[k-1];
           }

           for(int i=0;i<=end+1;i++)
           {
               if(!dp[i])
               {
                   cmax=i-1;
                   break;
               }
           }

           if(cmax>ans||(cmax==ans&&k<minnum)||(cmax==ans&&k==minnum&&a[k-1]<minmax))//三个条件
           {
               ans=cmax;
               minnum=k;
               minmax=a[k-1];
               memcpy(b,a,sizeof a);
           }
        }
      printf("max coverage = %3d :",ans);
        for(int i=0;i<minnum;i++)
         printf(" %2d",b[i]);
          cout<<endl;
    }
    return 0;
}