HDU 4499.Cannon 搜索
Cannon
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65535/65535 K (Java/Others)
Total Submission(s): 965 Accepted Submission(s): 556
An eat action, for example, Cannon A eating chessman B, requires two conditions:
1、A and B is in either the same row or the same column in the chess grid.
2、There is exactly one chessman between A and B.
Here comes the problem.
Given an N x M chess grid, with some existing chessmen on it, you need put maximum cannon pieces into the grid, satisfying that any two cannons are not able to eat each other. It is worth nothing that we only account the cannon pieces you put in the grid, and no two pieces shares the same cell.
In each test case, there are three positive integers N, M and Q (1<= N, M<=5, 0<=Q <= N x M) in the first line, indicating the row number, column number of the grid, and the number of the existing chessmen.
In the second line, there are Q pairs of integers. Each pair of integers X, Y indicates the row index and the column index of the piece. Row indexes are numbered from 0 to N-1, and column indexes are numbered from 0 to M-1. It guarantees no pieces share the same cell.
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,m;
int ans=;
int edge[][];
int dfs(int x,int y,int cou)
{
int i,j,t,sign,ok;
/*
for(i=0; i<n; i++)
{
for(j=0; j<m; j++)
cout<<edge[i][j]<<" ";
cout<<endl;
}
cout<<endl;
*/
if(cou>ans) ans=cou;
edge[x][y]=;
/**当前行可以增加*/
for(j=y+; j<n; j++)
{
if(edge[x][j]==) continue;
sign=;
ok=;
for(t=j-; t>=; t--)
{
if(edge[x][t]!=) sign++;
if(sign==&&edge[x][t]==)
{
ok=;
break;
}
}
if(ok==)
{
sign=;
for(t=x-; t>=; t--)
{
if(edge[t][j]!=) sign++;
if(sign==&&edge[t][j]==)
{
ok=;
break;
}
}
if(ok==)
{
edge[x][j]=;
dfs(x,j,cou+);
edge[x][j]=;
}
}
}
/**当前行不能增加,加入后面的行*/
for(i=x+; i<n; i++)
{
for(j=; j<m; j++)
{
if(edge[i][j]==) continue;
sign=;
ok=;
for(t=i-; t>=; t--)
{
if(edge[t][j]!=) sign++;
if(sign==&&edge[t][j]==)
{
ok=;
break;
}
}
if(ok==)
{
edge[i][j]=;
dfs(i,j,cou+);
edge[i][j]=;
}
}
}
}
int main()
{
int i,j,q;
int x,y;
while(scanf("%d%d%d",&n,&m,&q)!=EOF)
{
memset(edge,,sizeof(edge));
while(q--)
{
scanf("%d%d",&x,&y);
edge[x][y]=;
}
ans=;
for(i=; i<n; i++)
for(j=; j<m; j++)
if(edge[i][j]==)
{
edge[i][j]=;
dfs(i,j,);
edge[i][j]=;
}
cout<<ans<<endl;
}
return ;
}
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