Time Limit: 2000MS   Memory Limit: 32768KB   64bit IO Format: %lld & %llu

Submit
Status

Description

You are given a 2D board where in some cells there are gold. You want to fill the board with
2 x 1 dominoes such that all gold are covered. You may use the dominoes vertically or horizontally and the dominoes may overlap. All you have to do is to cover the gold with least number of dominoes.

In the picture, the golden cells denote that the cells contain gold, and the blue ones denote the
2 x 1 dominoes. The dominoes may overlap, as we already said, as shown in the picture. In reality the dominoes will cover the full
2 x 1 cells; we showed small dominoes just to show how to cover the gold with 11 dominoes.

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

Each case starts with a row containing two integers m (1 ≤ m ≤ 20) and
n (1 ≤ n ≤ 20) and m * n > 1. Here m represents the number of rows, and
n represents the number of columns. Then there will be m lines, each containing
n characters from the set ['*','o']. A
'*'
character symbolizes the cells which contains a gold, whereas an
'o'
character represents empty cells.

Output

For each case print the case number and the minimum number of dominoes necessary to cover all gold ('*' entries) in the given board.

Sample Input

2

5 8

oo**oooo

*oo*ooo*

******oo

*o*oo*oo

******oo

3 4

**oo

**oo

*oo*

Sample Output

Case 1: 11

Case 2: 4

Source

Problem Setter: Jane Alam Jan

题意:有n*m的图,*表示金矿所在地,要用2*1或者1*2的骨牌将金矿全部覆盖,问最少需要多少骨牌

将所有的金矿编号,根据他们的坐标和,坐标和为奇数跟偶数的分别编号,并记录相应的个数oddnum和evennum,因为骨牌只有2*1跟1*2的,这也就是说一个金矿只能跟相邻的金矿相连,现在就从坐标和为奇数的点开始向坐标和为偶数的点建边,求出最大匹配数,金矿总数减去最大匹配数就是要求的

#include<cstdio>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
#define MAXN 500
char s[25][25];
int used[MAXN],pipei[MAXN];
int map[25][25],m,n,oddnum,evennum,k=1;
vector<int>G[MAXN];
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
void getmap()
{
memset(map,0,sizeof(0));
for(int i=0;i<500;i++)
G[i].clear();
oddnum=evennum=0;
for(int i=0;i<n;i++)
for(int j=0;j<m;j++)
{
map[i][j]=-1;
if(s[i][j]=='*')
{
if((i+j)&1)
map[i][j]=++oddnum;
else
map[i][j]=++evennum;
}
}
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(s[i][j]=='*'&&((i+j)&1))
{
for(int k=0;k<4;k++)
{
int x=i+dx[k];
int y=j+dy[k];
if(x<0||x>=n||y<0||y>=m)
continue;
if(s[x][y]=='*')
G[map[i][j]].push_back(map[x][y]);
}
}
}
}
}
int find(int x)
{
for(int i=0;i<G[x].size();i++)
{
int y=G[x][i];
if(!used[y])
{
used[y]=1;
if(pipei[y]==-1||find(pipei[y]))
{
pipei[y]=x;
return 1;
}
}
}
return 0;
}
void solve()
{
memset(pipei,-1,sizeof(pipei));
int sum=0;
for(int i=1;i<=oddnum;i++)
{
memset(used,0,sizeof(used));
sum+=find(i);
}
printf("Case %d: %d\n",k++,oddnum+evennum-sum);
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
memset(s,'\0',sizeof(s));
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%s",s[i]);
getmap();
solve();
}
return 0;
}

LightOJ--1152--Hiding Gold(二分图奇偶建图)(好题)的更多相关文章

  1. 4185 Oil Skimming 最大匹配 奇偶建图

    题目大意: 统计相邻(上下左右)的‘#’的对数. 解法: 与题目hdu1507 Uncle Tom's Inherited Land*类似,需要用奇偶建图.就是行+列为奇数的作为X集合,偶尔作为Y集合 ...

  2. hdoj--5093--Battle ships(二分图经典建图)

    Battle ships Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Tot ...

  3. Battle ships(二分图,建图,好题)

    Battle ships Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Tot ...

  4. joj 2453 candy 网络流建图的题

    Problem D: Candy As a teacher of a kindergarten, you have many things to do during a day, one of whi ...

  5. poj 3281 Dining 网络流-最大流-建图的题

    题意很简单:JOHN是一个农场主养了一些奶牛,神奇的是这些个奶牛有不同的品味,只喜欢吃某些食物,喝某些饮料,傻傻的John做了很多食物和饮料,但她不知道可以最多喂饱多少牛,(喂饱当然是有吃有喝才会饱) ...

  6. POJ 2195 一人一房 最小费用流 建图 水题

    Going Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 21010   Accepted: 10614 Desc ...

  7. light oj 1152 Hiding Gold

    题目: You are given a 2D board where in some cells there are gold. You want to fill the board with 2 x ...

  8. [SCOI2007]修车(建图好题)

    [SCOI2007]修车 https://www.lydsy.com/JudgeOnline/problem.php?id=1070 Time Limit: 1 Sec  Memory Limit:  ...

  9. HDU1853 & 蜜汁建图+KM模板

    题意: 给你一个N个点M条边的带权有向图,现在要你求这样一个值:该有向图中的所有顶点正好被1个或多个不相交的有向环覆盖.这个值就是 所有这些有向环的权值和. 要求该值越小越好. SOL: 本来还想ta ...

随机推荐

  1. MEF编程模型

    Contract由Contract name和Contract type组成,Contract两个参数可以省略可以implicit也可以explicit,implicit时name和type会自动推断 ...

  2. mvc3结合spring.net-依赖注入

    namespace Tuzi.Models.IService { public interface IPersonService { string say(string words); } names ...

  3. CSS的常用属性(二)

    盒子模型之边框 border-(top/bottom/left/right)-style: solid 边框的风格 如(solid 实线,dotted 点线,dashed 虚线) border-top ...

  4. Excel的用到的常规的技巧

    这几天在做各种发票的报表,好几百的数据当然离不开EXCel,自己又是个白班,就记录下啦! EXCEL 判断某一单元格值是否包含在某一列中 就在Excel的表格中加入这个函数:=IF(ISERROR(V ...

  5. 【sqli-labs】 less17 POST - Update Query- Error Based - String (基于错误的更新查询POST注入)

    这是一个重置密码界面,查看源码可以看到username作了防注入处理 逻辑是先通过用户名查出数据,在进行密码的update操作 所以要先知道用户名,实际情况中可以注册用户然后实行攻击,这里先用admi ...

  6. vmware Horizon 7 与远程桌面(mstsc)兼容性问题解决办法

    关于Horizon 7 Agent与远程桌面(mstsc)兼容性问题解决办法 在Horizon 7环境中,在桌面模板安装了Horizon Agent后,就无法直接通过微软的远程桌面(mstsc)工具连 ...

  7. 用电销机器人让电销企业迈入AI智能时代

    2019年是AI智能快速发展的一年,有非常多的企业已经用AI智能代替原有的传统员工做重复性高的工作,就拿销售行业来说,为了让电销员工提升工作效率,拥有更多的成单,大部分有电销岗位的公司都会把重复率较高 ...

  8. tomcat 热加载设置

    找到tomcat项目的apache-tomcat-8.0.30\conf\context.xml,打开进行编辑,把Context项中加上 reloadable="true" < ...

  9. ADB 常用命令学习

    参考文档:https://www.cnblogs.com/bravesnail/articles/5850335.html非常感谢作者的分享,以下是我学习的记录.Android 常用adb 命令汇总- ...

  10. C#第十一节课

    类 using System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Thr ...