二分图匹配：
分别按行和列把图展开。hungary二分图匹配。
。。。
例子： 4 4 *ooo o### **#* ooo* 按行展开。
。。
。 *ooo o#oo oo#o ooo# **#o ooo* ooo* 再按列展开。。。
7 * 8 *ooooooo oooooooo oooooooo oooooooo *o*ooooo ooooooo* ooooooo* 匹配结果3
Battle ships
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)Total Submission(s): 106 Accepted Submission(s): 53
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.
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.
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
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=2600;
char mp[55][55];
char hang[maxn][maxn];
char lie[maxn][maxn];
int n,m;
int nn,mm;
void getHang()
{
nn=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(mp[i][j]=='o')
{
hang[nn][j]='o';
}
else if(mp[i][j]=='*')
{
hang[nn][j]='*';
}
else if(mp[i][j]=='#')
{
hang[nn][j]='#';
for(int k=j+1;k<m;k++)
{
hang[nn][k]='o';
}
if(j!=m-1)
{
nn++;
for(int k=0;k<=j;k++)
{
hang[nn][k]='o';
}
}
}
}
nn++;
}
/*********************debug***********************
cout<<" debug hang \n";
for(int i=0;i<nn;i++)
{
printf("%s\n",hang[i]);
}
*********************debug***********************/
}
void getLie()
{
mm=0;
for(int i=0;i<m;i++)
{
for(int j=0;j<nn;j++)
{
if(hang[j][i]=='*')
{
lie[j][mm]='*';
}
else if(hang[j][i]=='o')
{
lie[j][mm]='o';
}
else if(hang[j][i]=='#')
{
for(int k=j;k<nn;k++)
{
lie[k][mm]='o';
}
if(j!=nn-1)
{
mm++;
for(int k=0;k<=j;k++)
{
lie[k][mm]='o';
}
}
}
}
mm++;
}
/**************debug lie*******************
cout<<"debug lie\n";
cout<<nn<<" * "<<mm<<endl;
for(int i=0;i<nn;i++)
{
for(int j=0;j<mm;j++)
{
printf("%c",lie[i][j]);
if(j==mm-1) putchar(10);
}
}
**************debug lie*******************/
}
struct Edge
{
int to,next;
}edge[maxn*maxn];
int Adj[maxn],Size;
void init()
{
memset(Adj,-1,sizeof(Adj)); Size=0;
}
void add_edge(int u,int v)
{
edge[Size].to=v; edge[Size].next=Adj[u]; Adj[u]=Size++;
}
int linker[maxn];
bool used[maxn];
bool dfs(int u)
{
for(int i=Adj[u];~i;i=edge[i].next)
{
int v=edge[i].to;
if(!used[v])
{
used[v]=true;
if(linker[v]==-1||dfs(linker[v]))
{
linker[v]=u;
return true;
}
}
}
return false;
}
int hungary()
{
int res=0;
memset(linker,-1,sizeof(linker));
for(int u=0;u<nn;u++)
{
memset(used,false,sizeof(used));
if(dfs(u)) res++;
}
return res;
}
int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
memset(mp,0,sizeof(mp));
memset(hang,0,sizeof(hang));
memset(lie,0,sizeof(lie));
init();
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
scanf("%s",mp[i]);
getHang();
getLie();
/// build graph
for(int i=0;i<nn;i++)
{
for(int j=0;j<mm;j++)
{
if(lie[i][j]=='*')
{
add_edge(i,j);
}
}
}
printf("%d\n",hungary());
}
return 0;
}
本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5262022.html，如需转载请自行联系原作者
