枚举角度DFS。。。。
Lines
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 479 Accepted Submission(s): 140
Problem Description
You play a game with your friend. He draws several lines on the paper with n×m square grids (see the left figure). After that, he writes down the number of lines passing through every integer coordinate in a matrix (see the right figure).
The number of lines passing though coordinate (i,j) is written in cell (i,j) in the right figure.(i,j both start from 0).
The number of lines passing though coordinate (i,j) is written in cell (i,j) in the right figure.(i,j both start from 0).
You are given the matrix written by your friend. You need to figure out the possible minimal number of lines your friend drew on the paper.
Input
The first line of the input contains an integer T indicating the number of test cases( 0 < T <= 10).
For each test case, the first line contains two integers n, m (1 ≤ n, m ≤ 50) representing the size of the grids on the paper. The following (n+1) × (m+1) numbers is what your friend writes. It is guaranteed that the number of lines your friend draws does not
exceed 14. Each line passes through integer coordinates at least three times.
Output
For each test case, you need to output the minimal number of lines your friend drew on the paper in a single line.
Sample Input
1 5 4 0 1 0 0 1 0 1 0 1 0 2 1 1 0 0 0 3 1 0 0 1 1 1 0 1 0 1 0 1 0
Sample Output
4
Source
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <map>
using namespace std;
map<double,bool> DB;
int n,m,ans;
int mp[60][60];
inline bool is_in(int x,int y)
{
if(x>=0&&x<n&&y>=0&y<m) return true;
return false;
}
void dfs(int deep,int remain)
{
if(deep>=ans) return ;
if(remain==0)
{
ans=min(ans,deep);
return ;
}
bool flag=false;
for(int i=0;i<n&&!flag;i++)
{
for(int j=0;j<m&&!flag;j++)
{
if(mp[i][j])
{
flag=true;
if(i==0)
{
int jian=0;
for(int p=0;p<n;p++)
{
if(mp[p][j]) jian++;
else break;
}
if(jian==n)
{
for(int p=0;p<n;p++) mp[p][j]--;
dfs(deep+1,remain-n);
for(int p=0;p<n;p++) mp[p][j]++;
}
}
if(j==0)
{
int jian=0;
for(int q=0;q<m;q++)
{
if(mp[i][q]) jian++;
else break;
}
if(jian==m)
{
for(int q=0;q<m;q++) mp[i][q]--;
dfs(deep+1,remain-m);
for(int q=0;q<m;q++) mp[i][q]++;
}
}
DB.clear();
for(int p=i+1;p<n;p++)
{
for(int q=0;q<m;q++)
{
if(mp[p][q]&&q!=j)
{
int x=p-i,y=q-j;
double slope=y*1./x;
if(DB[slope]==true) continue;
DB[slope]=true;
int cnt=0;
while(is_in(i+cnt*x,j+cnt*y)&&mp[i+cnt*x][j+cnt*y]) cnt++;
if(is_in(i-x,j-y)) continue;
if(is_in(i+cnt*x,j+cnt*y)) continue;
if(cnt<3) continue;
for(int c=0;c<cnt;c++) mp[i+c*x][j+c*y]--;
dfs(deep+1,remain-cnt);
for(int c=0;c<cnt;c++) mp[i+c*x][j+c*y]++;
}
}
}
}
}
}
}
int main()
{
int T_T;
scanf("%d",&T_T);
while(T_T--)
{
scanf("%d%d",&n,&m);
n++; m++; int sum=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
scanf("%d",&mp[i][j]);
sum+=mp[i][j];
}
}
ans=min(14,sum/3);
dfs(0,sum);
printf("%d\n",ans);
}
return 0;
}