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  • hdu 1498 50 years, 50 colors(二分匹配_匈牙利算法)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1498

    50 years, 50 colors

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1918    Accepted Submission(s): 1058


    Problem Description
    On Octorber 21st, HDU 50-year-celebration, 50-color balloons floating around the campus, it's so nice, isn't it? To celebrate this meaningful day, the ACM team of HDU hold some fuuny games. Especially, there will be a game named "crashing color balloons".

    There will be a n*n matrix board on the ground, and each grid will have a color balloon in it.And the color of the ballon will be in the range of [1, 50].After the referee shouts "go!",you can begin to crash the balloons.Every time you can only choose one kind of balloon to crash, we define that the two balloons with the same color belong to the same kind.What's more, each time you can only choose a single row or column of balloon, and crash the balloons that with the color you had chosen. Of course, a lot of students are waiting to play this game, so we just give every student k times to crash the balloons.

    Here comes the problem: which kind of balloon is impossible to be all crashed by a student in k times.

     
    Input
    There will be multiple input cases.Each test case begins with two integers n, k. n is the number of rows and columns of the balloons (1 <= n <= 100), and k is the times that ginving to each student(0 < k <= n).Follow a matrix A of n*n, where Aij denote the color of the ballon in the i row, j column.Input ends with n = k = 0.
     
    Output
    For each test case, print in ascending order all the colors of which are impossible to be crashed by a student in k times. If there is no choice, print "-1".
     
    Sample Input
    1 1
    1
    2 1
    1 1
    1 2
    2 1
    1 2
    2 2
    5 4
    1 2 3 4 5
    2 3 4 5 1
    3 4 5 1 2
    4 5 1 2 3
    5 1 2 3 4
    3 3
    50 50 50
    50 50 50
    50 50 50
    0 0
     
    Sample Output
    -1
    1
    2
    1 2 3 4 5
    -1
     
    Author
    8600
     
    Source
     
     
    题目大意:刷气球,气球有很多种颜色。 每次刷气球可以刷一整行或者一整列,刷m次,最后输出刷不完的气球数字。(每个不同的数字代表不同颜色的气球)。
    特别注意:气球的颜色数字范围在1~50之间。
     
    详见代码(代码解释)
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 
     5 using namespace std;
     6 
     7 int n,m;
     8 int vis[110],Map[110][110],ok[110],cc[55];
     9 
    10 bool Find(int x,int now)
    11 {
    12     for (int i=1; i<=n; i++)
    13     {
    14         if (!vis[i]&&Map[x][i]==now)
    15         {
    16             vis[i]=1;
    17             if (!ok[i])
    18             {
    19                 ok[i]=x;
    20                 return true;
    21             }
    22             else
    23             {
    24                 if (Find(ok[i],now))
    25                 {
    26                     ok[i]=x;
    27                     return true;
    28                 }
    29             }
    30         }
    31     }
    32     return false;
    33 }
    34 
    35 int main()
    36 {
    37     int flag,ans,s[110];
    38     while (~scanf("%d%d",&n,&m))
    39     {
    40         flag=0;
    41         memset(cc,0,sizeof(cc));
    42         memset(Map,0,sizeof(Map));
    43         if (n==0&&m==0)
    44             break;
    45         for (int i=1; i<=n; i++)
    46         {
    47             for (int j=1; j<=n; j++)
    48             {
    49                 scanf("%d",&Map[i][j]);
    50                 cc[Map[i][j]]=1;//记录这个颜色出现过
    51             }
    52         }
    53         memset(ok,0,sizeof(ok));
    54         int k=0;
    55         for (int i=1; i<=50; i++)
    56         {
    57             memset(ok,0,sizeof(ok));
    58             ans=0;
    59             if (cc[i]==1)    //判断颜色是否出现过,加入这里不进行标记的话最后输出的是没有被刷完的颜色数字,就会输出一些乱七八糟的东西
    60             {
    61                 for (int j=1; j<=n; j++)
    62                 {
    63                     memset(vis,0,sizeof(vis));
    64                     if (Find(j,i))
    65                         ans++;
    66                 }
    67                 if (ans>m)   //如果最大匹配数大于m次的话就是刷不完的
    68                 {
    69                     s[++k]=i;
    70                     flag=1;
    71                 }
    72             }
    73         }
    74         if (flag==0)
    75             printf ("-1
    ");
    76         else
    77         {
    78             for (int i=1; i<k; i++)    //控制输出格式
    79             {
    80                 printf ("%d ",s[i]);
    81             }
    82             printf ("%d
    ",s[k]);
    83         }
    84     }
    85     return 0;
    86 }
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  • 原文地址:https://www.cnblogs.com/qq-star/p/4707383.html
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