Battle ships
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 589 Accepted Submission(s): 233
Problem Description
Dear contestant, now you are an excellent navy commander, who is responsible of a tough mission currently.
Your fleet unfortunately encountered an enemy fleet near the South Pole where the geographical conditions are negative for both sides. The floating ice and iceberg blocks battleships move which leads to this unexpected engagement highly dangerous, unpredictable
and incontrollable.
But, fortunately, as an experienced navy commander, you are able to take opportunity to embattle the ships to maximize the utility of cannons on the battleships before the engagement.
The target is, arrange as many battleships as you can in the map. However, there are three rules so that you cannot do that arbitrary:
A battleship cannot lay on floating ice
A battleship cannot be placed on an iceberg
Two battleships cannot be arranged in the same row or column, unless one or more icebergs are in the middle of them.
Input
There is only one integer T (0<T<12) at the beginning line, which means following T test cases.
For each test case, two integers m and n (1 <= m, n <= 50) are at the first line, represents the number of rows and columns of the battlefield map respectively. Following m lines contains n characters iteratively, each character belongs to one of ‘#’, ‘*’,
‘o’, that symbolize iceberg, ordinary sea and floating ice.
Output
For each case, output just one line, contains a single integer which represents the maximal possible number of battleships can be arranged.
Sample Input
2 4 4 *ooo o### **#* ooo* 4 4 #*** *#** **#* ooo#
Sample Output
3 5
链接：http://acm.hdu.edu.cn/showproblem.php?pid=5093
题意：选尽可能多*号点，要求选得点之间必须有#阻隔，否者不能在同一行或者同一列。
做法：
先要把竖向*和O相连的 都连成一起，并编号。
然后是横着的也一样。
案例编号后：
2
4 4
*ooo
o###
**#*
ooo*
这里是竖向编号，同编号为一块
1 2 4 6
1 0 0 0
1 3 0 7
1 3 5 7
这里是横向编号
1 1 1 1
2 0 0 0
3 3 0 4
5 5 5 5
3
4 4
#***
*#**
**#*
ooo#
1 2 4 6
1 0 4 6
1 3 0 6
1 3 5 7
1 1 1 1
2 0 3 3
4 4 0 5
6 6 6 5
5
然后把竖向的每一块 都看作二分图左边的每一个点，
然后把横向的每一块 都看作二分图右边的每一个点，
星号必会对应一个竖向的块 编号， 和一个横向的块编号。
把每个星号对应的 两个编号连线。
如果匹配了，就表示这两个块都被占据了，这两个块就不能再有其他星号点被连了。
所以要知道最多有多少*可以放，就只用知道最多可以有几条边可以选。也就是求最大二分匹配了。
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <malloc.h>
#include <ctype.h>
#include <math.h>
#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
#include <stack>
#include <queue>
#include <vector>
#include <deque>
#include <set>
#include <map>
#define N 3000
int visit[N];
int mark[N];
int match[N][N];
int n,m,k;
int dfs(int x)
{
int i;
//printf("%d\n",x);
for(i=1;i<=m;i++)//对左边的节点x与右边的节点进行逐一检查
{
if(!visit[i]&&match[x][i])
{
visit[i]=1;//标记检查过的点
if(mark[i]==-1||dfs(mark[i])) //mark如果没男的，就直接给个老公，如果已经有老公，搜他老公有没有喜欢的，有的话，该女配新男；；；
{//|| 前面过了 后面不运行；；
mark[i]=x;//修改匹配关系
return 1;
}
}
}
return 0;
}
int hungary ()
{
memset(mark,-1,sizeof(mark));
int maxx=0,j;
for(j=1;j<=n;j++)//对做部分顶点逐个进行遍历
{
memset(visit,0,sizeof(visit));
if(dfs(j))
maxx++;
}
return maxx;
}
char mp[60][60];
int cdh[60][60];
int cds[60][60];
int main()//注意 mark[m]=n mark[j] = i
{
int maxx;
int nn,mm;
int t;
scanf("%d",&t);
while(t--) //k 个配
{
scanf("%d%d",&nn,&mm);
/*
for(int i=0;i<nn;i++)
{
for(int j=0;j<mm;j++)
{
if((i+j)%2==1)
mp[i][j]='#';
else
mp[i][j]='*';
}
}*/
for(int i=0;i<nn;i++)
{
scanf("%s",mp[i]);
}
int sqi=0;
for(int j=0;j<mm;j++)//左竖着
{
for(int i=0;i<nn;i++)
{
if(mp[i][j]!='#')
{
if(i==0)
cds[i][j]=++sqi;
else if(mp[i-1][j]=='#')
cds[i][j]=++sqi;
else
cds[i][j]=sqi;
}
}
}
n=sqi;
int hqi=0;
for(int i=0;i<nn;i++) //you heng 着
{
for(int j=0;j<mm;j++)
{
if(mp[i][j]!='#')
{
if(j==0)
cdh[i][j]=++hqi;
else if(mp[i][j-1]=='#')
cdh[i][j]=++hqi;
else
cdh[i][j]=hqi;
}
}
}
m=hqi;
/*
for(int i=0;i<nn;i++)//左竖着
{
for(int j=0;j<mm;j++)
{
printf("%d ",cdh[i][j]);
}
puts("");
}
*/
memset(match,0,sizeof(match));
for(int i=0;i<nn;i++)
{
for(int j=0;j<mm;j++)
{
if(mp[i][j]=='*')
{
int z=cds[i][j];
int y=cdh[i][j];
match[z][y]=1;
}
}
}
maxx=hungary();
printf ("%d\n",maxx);
/*
scanf ("%d%d%d",&n,&m,&k)!=EOF
for(i=1;i<=k;i++)
{
int a,b;
scanf ("%d%d",&a,&b);
match[a][b]=1;//标记当前匹配关系
}
max=hungary();
printf ("%d\n",max);*/
}
return 0;
}