zoukankan      html  css  js  c++  java
  • hdu 1498 50 years, 50 colors

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 2717    Accepted Submission(s): 1551

    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
     
    Sample Output
     
    Author
    8600
     
     
    二分图最大匹配
    匈牙利算法
    卡输出格式 坑!
    #include <algorithm>
    #include <cstring>
    #include <cstdio>
    #define N 105
    
    using namespace std;
    bool vis[N],used[N];
    int G[N][N],a[N][N],n,K,f[N],ans[N],sum;
    inline int init()
    {
        sum=0;
        memset(vis,0,sizeof(vis));
    }
    bool dfs(int x)
    {
        for(int i=1;i<=n;++i)
        {
            if(!used[i]&&G[x][i])
            {
                used[i]=1;
                if(f[i]==-1||dfs(f[i]))
                {
                    f[i]=x;
                    return 1;
                }
            }
        }
        return 0;
    }
    int check()
    {
        int num=0;
        memset(f,-1,sizeof(f));
        for(int i=1;i<=n;++i)
        {
            memset(used,0,sizeof(used));
            if(dfs(i)) num++;
        }
        return num;
    }
    int Main()
    {
        for(;scanf("%d%d",&n,&K)&&n&&K;)
        {
            init();
            for(int i=1;i<=n;++i)
             for(int j=1;j<=n;++j)
              scanf("%d",&a[i][j]);
            for(int i=1;i<=n;++i)
            {
                for(int j=1;j<=n;++j)
                {
                    if(vis[a[i][j]]) continue;
                    vis[a[i][j]]=1;
                    memset(G,0,sizeof(G));
                    for(int k=1;k<=n;++k)
                     for(int l=1;l<=n;++l)
                      if(a[k][l]==a[i][j]) G[k][l]=1;
                    if(check()>K) ans[++sum]=a[i][j]; 
                }
            }
            if(!sum) {printf("-1
    ");continue;}
            sort(ans+1,ans+1+sum);
            for(int i=1;i<sum;++i) printf("%d ",ans[i]);
            printf("%d
    ",ans[sum]); 
        }
        return 0;
    }
    int sb=Main();
    int main(int argc,char *argv[]){;}
    我们都在命运之湖上荡舟划桨,波浪起伏着而我们无法逃脱孤航。但是假使我们迷失了方向,波浪将指引我们穿越另一天的曙光。
  • 相关阅读:
    MacOS更改zsh命令行前缀
    python中os._exit()和sys.exit(), exit(0)和exit(1) 的用法和区别
    如何解析 redis 的 rdb 文件
    流量回放工具<二>
    策略路由配置<一>
    h3c镜像模式配置
    python上传gz文件请求
    优先队列(大顶堆实现)
    bm和kmp和bf
    dedecms 软件下载模块加入flashget快车下载代码
  • 原文地址:https://www.cnblogs.com/ruojisun/p/7543370.html
Copyright © 2011-2022 走看看