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其实嘞，这个线可以只延伸一端

然后嘞，爆搜一次就可以

最后嘞，600-800ms过

本弱就是弱啊，你来打我呀……

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
int a[100][100];
int n,m,ans;
bool dfs(int step)
{
	int i,j,t,ii,jj,x,y,cnt,tx,ty;
	t=0;
	for(i=0;i<n;i++)
		t=max(t,*max_element(a[i],a[i]+m));
	if(step+t>=ans)
		return 0;
	if(t==0)
	{
		ans=step;
		return 1;
	}
	for(i=0;i<n;i++)
		for(j=0;j<m;j++)
		{
			if(a[i][j])
			{
				if(i==0)
				{
					for(ii=0;ii<n;ii++)
						if(!a[ii][j])
							break;
					if(ii==n)
					{
						for(ii=0;ii<n;ii++)
							a[ii][j]--;
						if(dfs(step+1))
							return 1;
						for(ii=0;ii<n;ii++)
							a[ii][j]++;
					}
				}
				if(j==0)
				{
					for(ii=0;ii<m;ii++)
						if(!a[i][ii])
							break;
					if(ii==m)
					{
						for(ii=0;ii<m;ii++)
							a[i][ii]--;
						if(dfs(step+1))
							return 1;
						for(ii=0;ii<m;ii++)
							a[i][ii]++;
					}
				}
				for(ii=i+1;ii<n;ii++)
					for(jj=0;jj<m;jj++)
					{
						if(a[ii][jj]&&jj!=j)
						{
							x=ii-i;
							y=jj-j;
							cnt=0;
							for(tx=i,ty=j;tx>=0&&tx<n&&ty>=0&&ty<m&&a[tx][ty];tx+=x,ty+=y)
								cnt++;
							if(tx>=0&&tx<n&&ty>=0&&ty<m||cnt<3)
								continue;
							for(tx=i,ty=j;tx>=0&&tx<n&&ty>=0&&ty<m&&a[tx][ty];tx+=x,ty+=y)
								a[tx][ty]--;
							if(dfs(step+1))
								return 1;
							for(tx=i,ty=j;tx>=0&&tx<n&&ty>=0&&ty<m;tx+=x,ty+=y)
								a[tx][ty]++;
						}
					}
				return 0;
			}
		}
}
int main()
{
	int T,i,j,cnt;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%d",&n,&m);
		n++;m++;
		cnt=0;
		for(i=0;i<n;i++)
			for(j=0;j<m;j++)
			{
				scanf("%d",&a[i][j]);
				cnt+=a[i][j];
			}
		ans=min(14,cnt/3);
		dfs(0);
		printf("%d\n",ans);
	}
}

Lines

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).

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.

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

4

hdu 5031 Lines 爆搜

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原文地址：http://blog.csdn.net/stl112514/article/details/41048207