UVALive-7470-Easy Problemset（模拟）

Perhaps one of the hardest problems of any ACM ICPC contest is to create a problemset with a
reasonable number of easy problems. On Not Easy European Regional Contest this problem is solved
as follows.
There are n jury members (judges). They are numbered from 1 to n. Judge number i had prepared
pi easy problems before the jury meeting. Each of these problems has a hardness between 0 and 49 (the
higher the harder). Each judge also knows a very large (say infinite) number of hard problems (their
hardness is 50). Judges need to select k problems to be used on the contest during this meeting.
They start to propose problems in the ascending order of judges numbers. The first judge takes the
first problem from his list of remaining easy problems (or a hard problem, if he has already proposed all
his easy problems) and proposes it. The proposed problem is selected for the contest if its hardness
is greater than or equal to the total hardness of the problems selected so far, otherwise it
is considered too easy. Then the second judge does the same etc.; after the n-th judge, the first one
proposes his next problem, and so on. This procedure is stopped immediately when k problems are
selected.
If all judges have proposed all their easy problems, but they still have selected less than k problems,
then they take some hard problems to complete the problemset regardless of the total hardness.
Your task is to calculate the total hardness of the problemset created by the judges.
Input
The input file contains several test cases, each of them as described below.
The first line of the input file contains the number of judges n (2 ≤ n ≤ 10) and the number of
problems k (8 ≤ k ≤ 14). The i-th of the following n lines contains the description of the problems
prepared by the i-th judge. It starts with pi (1 ≤ pi ≤ 10) followed by pi non negative integers between
0 and 49 — the hardnesses of the problems prepared by the i-th judge in the order they will be proposed.
Output
For each test case, output a only integer — the total hardness of the selected problems — on a line by
itself.
Note:
In the first example, three problems with hardnesses of 0, 1, and 1 are selected first. Then the
first judge proposes the problem with hardness 3 and it is selected, too. The problem proposed by
the second judge with hardness 1 is not selected, because it is too easy. Then the problems proposed
by the third, the first, and the second judges are selected (their hardnesses are 5, 12 and 23). The
following three proposed problems with hardness of 17, 1 and 20 are not selected, and the problemset
is completed with a problem proposed by the third judge with hardness of 49. So the total hardness of
the problemset is 94.
In the second example, three problems with hardnesses of 1, 1, and 2 are selected first. The second
problem of the first judge (hardness 3) is too easy. The second judge is out of his easy problems, so he
proposes a problem with hardness 50 and it is selected. The third judge’s problem with hardness 5 is
not selected. The judges decide to take 6 more hard problems to complete the problemset, which gives
the total hardness of 54 + 6·50 = 354.
Sample Input
3 8
5 0 3 12 1 10
4 1 1 23 20
4 1 5 17 49
3 10
2 1 3
1 1
2 2 5
Sample Output
94
354

题意：多组数据，第一行给出N，K，代表接着有N行，总共需要选取K个数。接着N行每行第一个数字为P，代表后面有P个数字。接着从上往下从左往右选数，只有当所选数总和小于等于待选数时，才可以选择待选数，如果当前位置空着，那么就把当前位置数值看成50，如果最后数不够选，那么全部用50补齐。输出最终选择数的总和

代码

#include<iostream>#include<algorithm>#include<stdio.h>#include<string.h>#include<queue>#include<string.h>#include<math.h>using namespace std;const int maxn=15;int num[maxn][maxn];int main(){    int N;//行数    int K;    while(~scanf("%d%d",&N,&K))    {        memset(num,-1,sizeof(num));        int max_p=22;//最大列数        for(int i=1; i<=N; i++) //建图        {            int P;            scanf("%d",&P);            max_p=max(max_p,P);            for(int j=1; j<=P; j++)                scanf("%d",&num[i][j]);        }        int num_k=0;//已经找到的数量        int sum=0;//已经找到的总和        for(int j=1; j<=max_p&&num_k!=K; j++)        {            for(int i=1; i<=N&&num_k!=K; i++)            {                if(num[i][j]!=-1)                {                    if(num[i][j]>=sum)                    {                        num_k++;                        sum+=num[i][j];                    }                }                else                {                    if(sum<=50)                    {                        num_k++;                        sum+=50;                        break;                    }                }                if(num_k==K)                    break;            }        }        if(K>num_k)            sum=sum+(K-num_k)*50;        printf("%d\n",sum);    }    return 0;}