校省选赛第一场A题Cinema题解

今天是学校省选的第一场比赛，0战绩收工，死死啃着A题来做，偏偏一直WA在TES1。

赛后，才发现，原来要freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);……

后来交了一下，Accepted了……烦恼！！！！！当是一个教训吧。

题目

Description
input
input.txt
output
output.txt

Overall there are m actors in Berland. Each actor has a personal identifier — an integer from 1 to m (distinct actors have distinct identifiers). Vasya likes to watch Berland movies with Berland actors, and he has k favorite actors. He watched the movie trailers for the next month and wrote the following information for every movie: the movie title, the number of actors who starred in it, and the identifiers of these actors. Besides, he managed to copy the movie titles and how many actors starred there, but he didn't manage to write down the identifiers of some actors. Vasya looks at his records and wonders which movies may be his favourite, and which ones may not be. Once Vasya learns the exact cast of all movies, his favorite movies will be determined as follows: a movie becomes favorite movie, if no other movie from Vasya's list has more favorite actors.

Help the boy to determine the following for each movie:
whether it surely will be his favourite movie;
whether it surely won't be his favourite movie;
can either be favourite or not.

Input

The first line of the input contains two integers m and k (1 ≤ m ≤ 100, 1 ≤ k ≤ m) — the number of actors in Berland and the number of Vasya's favourite actors.

The second line contains k distinct integers ai (1 ≤ ai ≤ m) — the identifiers of Vasya's favourite actors.

The third line contains a single integer n (1 ≤ n ≤ 100) — the number of movies in Vasya's list.

Then follow n blocks of lines, each block contains a movie's description. The i-th movie's description contains three lines:

    the first line contains string si (si consists of lowercase English letters and can have the length of from 1 to 10 characters, inclusive) — the movie's title, 

the second line contains a non-negative integer di (1 ≤ di ≤ m) — the number of actors who starred in this movie,
the third line has di integers bi, j (0 ≤ bi, j ≤ m) — the identifiers of the actors who star in this movie. If bi, j = 0, than Vasya doesn't remember the identifier of the j-th actor. It is guaranteed that the list of actors for a movie doesn't contain the same actors.

All movies have distinct names. The numbers on the lines are separated by single spaces.

Output

Print n lines in the output. In the i-th line print:

    0, if the i-th movie will surely be the favourite;
    1, if the i-th movie won't surely be the favourite;
    2, if the i-th movie can either be favourite, or not favourite. 

Sample Input
Input

5 3
1 2 3
6
firstfilm
3
0 0 0
secondfilm
4
0 0 4 5
thirdfilm
1
2
fourthfilm
1
5
fifthfilm
1
4
sixthfilm
2
1 0

Output

2
2
1
1
1
2

Input

5 3
1 3 5
4
jumanji
3
0 0 0
theeagle
5
1 2 3 4 0
matrix
3
2 4 0
sourcecode
2
2 4

Output

2
0
1
1

Hint

Note to the second sample:

    Movie jumanji can theoretically have from 1 to 3 Vasya's favourite actors.
    Movie theeagle has all three favourite actors, as the actor Vasya failed to remember, can only have identifier 5.
    Movie matrix can have exactly one favourite actor.
    Movie sourcecode doesn't have any favourite actors. 

Thus, movie theeagle will surely be favourite, movies matrix and sourcecode won't surely be favourite, and movie jumanji can be either favourite (if it has all three favourite actors), or not favourite.

看题就看得很纠结啦~不过总体来说大概明白。

思路就是：

1.如果一个电影最好的情况都比其他电影最差的情况差，那么该电影一定不是最好的，即状态为1

2.再判断如果一个电影不满足最差的情况都比其他电影最好的情况好，那么该电影不一定是最好的，即状态为2

3.最后剩余的就是最好的了。

特别绕啦~关键在于，一定不是最好的，不一定是最好的，略坑。

/*******************************************************************************/
/* OS           : 3.2.0-58-generic #88-Ubuntu SMP Tue Dec 3 UTC 2013 GNU/Linux
 * Compiler     : g++ (GCC)  4.6.3 (Ubuntu/Linaro 4.6.3-1ubuntu5)
 * Encoding     : UTF8
 * Date         : 2014-04-01
 * All Rights Reserved by yaolong.
*****************************************************************************/
/* Description: ***************************************************************
*****************************************************************************/
/* Analysis: ******************************************************************
*****************************************************************************/
/*****************************************************************************/

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
using namespace std;
class Movie
{
public:
    string name;
    int num;
    int stat;
    int like;
    int hate;
    int zero;
    vector<int> act;
    Movie() {};
    void resiz()
    {
        act.resize(num);
    }

};

int main()
{
    freopen("input.txt","r",stdin);

    freopen("output.txt","w",stdout);

 int cases=0;
    int m,k,mov_num,i,j,tmp;
    int fav[120];
    vector<Movie> mv;
    while(cin>>m>>k)
    {
        memset(fav,0,sizeof(fav));
        for(i=0; i<k; i++)
        {
            cin>>tmp;
            fav[tmp]=1;
        }
        cin>>mov_num;
        mv.clear();
        mv.resize(mov_num);
        for(i=0; i<mov_num; i++)
        {
            cin>>mv[i].name;
            cin>>mv[i].num;
            mv[i].resiz();
            for(j=0; j<mv[i].num; j++)
            {
                cin>>mv[i].act[j];
            }

        }
        int maxlike=-1;
        int minlike=-1;
        for(i=0; i<mov_num; i++)
        {
            mv[i].zero=0;
            mv[i].hate=0;
            mv[i].like=0;

            for(j=0; j<mv[i].act.size(); j++)
            {
                //统计0的个数
                if(mv[i].act[j]==0)
                {
                    mv[i].zero++;
                }
                else
                {
                    if(fav[mv[i].act[j]])
                    {
                        mv[i].like++;
                    }
                    else
                    {
                        mv[i]. hate++;
                    }
                }

            }

        }

        for(i=0; i<mov_num; i++)
        {

            int tmpbest=mv[i].like+min(mv[i].zero,k-mv[i].like);        //最好的情况

            int tmpworst=mv[i].like+max(0,mv[i].zero-(m-k-mv[i].hate)); //最坏的情况
            int cnt=0;

            for(j=0; j<mov_num; j++)
            {
                if(i!=j)
                {

                    if(tmpbest<mv[j].like+max(0,mv[j].zero-(m-k-mv[j].hate)))
                    {
                        mv[i].stat=1; //不满足最好的大于其他最小的，那么肯定不是最好的

                        break;
                    }

                }

            }
            for(j=0; j<mov_num&&mv[i].stat!=1; j++)
            {
                if(i!=j)
                {
                    if(tmpworst<mv[j].like+min(mv[j].zero,k-mv[j].like))
                    {
                        mv[i].stat=2    ; //不满足最差的大于其他最好的，那么这个就肯定不是最好的！同时又不是最差的，那么就是不确定的
                        break;
                    }

                }

            }
            mv[i].stat= mv[i].stat>=1?mv[i].stat:0; //如果没有被确定不是最好的或者未确定的，那么就是最右的

        }
        for(i=0; i<mov_num; i++)
        {
            cout<<mv[i].stat<<endl;

        }
    }

    return 0;

}