Two Circle:这道题的要求就是找到不想交的两个圆,使得在这两个圆内的数字累加和达到最大。并输出符合条件的圆的对数。
优化途径:(参考了keepit的代码)
(1)求出半径为r的圆,在每一行的宽度。
(2)对于table中的每一行i,求出最左端到当前列的累加和accum[i][j]。这样在计算圆内数字和的时候,不需一个一个累加,只需要将圆内每一行的最右端的累加和减去最左端的累加和(其实是最左端的左边一个位置),就可以得到园内每一行的数字累加。
(3)对于寻找与当前的圆不想交的另外一个圆:我们可以通过计算行偏移为xx的时候,列至少应该偏移多少。
(4)在(3)确定列偏移的情况下,我们需要寻找在往左偏移之后的左边的最大圆,以及在往右偏移之后的右边的最大圆。要想快速找到最大圆,我们可以事先对表中的每个位置(i,j)求出其第i行j左边的最大圆和j右边的最大圆,并保存最大圆个数的计数。
#include <iostream>
#include <vector>
#include <cmath>
using namespace std;
int n, m, r;
vector<vector<int> > cell(510, vector<int>(510, 0));
vector<vector<int> > accum(510, vector<int>(510, 0));
vector<vector<int> > circle(510, vector<int>(510,0));
vector<vector<pair<int,int> > > statL, statR;
vector<int> width;
vector<int> offset;
int Around(int row, int col)
{
/* Optimization method 2 */
int b = col - width[0];
int e = col + width[0];
int ans = accum[row][e] - accum[row][b-1];
for(int di = 1; di <= r; ++di)
{
b = col - width[di];
e = col + width[di];
ans += accum[row-di][e] - accum[row-di][b-1];
ans += accum[row+di][e] - accum[row+di][b-1];
}
return ans;
}
void Init()
{
/* Optimization method 1 */
width.clear();
width.resize( r + 1 );
int r2 = r*r;
for(int di = 0; di <= r; ++di)
{
width[di] = (int)sqrt(r2 - di*di);
}
/* Optimization method 3 */
int maxDi = r + r;
offset.clear();
offset.resize( maxDi + 1 );
for(int di = 0; di <= maxDi; ++di)
{
int maxD = min(di, r);
int maxOffset = 0;
for(int d = 0; d <= maxD; ++d)
{
if(di - d > r)
continue;
maxOffset = max(maxOffset, width[di-d]+width[d]+1);
}
offset[di] = maxOffset;
}
/* Optimization method 4 */
pair<int,int> val(0, 0);
statL.assign(n+1, vector<pair<int,int> >(m+5, val));
statR.assign(n+1, vector<pair<int,int> >(m+5, val));
for(int i = r + 1; i <= n - r; ++i)
{
for(int j = r + 1; j <= m - r; ++j)
{
circle[i][j] = Around(i, j);
statL[i][j] = statL[i][j-1];
int temp = statL[i][j].first;
if( circle[i][j] == temp)
++statL[i][j].second;
else if( circle[i][j] > temp )
statL[i][j] = make_pair(circle[i][j], 1);
}
}
for(int i = r + 1; i <= n - r; ++i)
{
for(int j = m - r; j >= r + 1; --j)
{
statR[i][j] = statR[i][j+1];
if( circle[i][j] == statR[i][j].first )
++statR[i][j].second;
else if( circle[i][j] > statR[i][j].first )
statR[i][j] = make_pair(circle[i][j], 1);
}
}
}
void Work()
{
/* Begin to solve the problem with the optimized results */
int ans = 0;
long long num = 0;
for(int i = r + 1; i <= n - r; ++i)
{
for(int j = r + 1; j <= m - r; ++j)
{
int temp = 0, temp_num = 0;
for(int ii = i; ii <= n - r; ++ii)
{
int di = ii - i;
int offsetVal = (di<offset.size()?offset[di]:0);
int jl = j - offsetVal;
int jr = j + offsetVal;
jl==jr?--jl:NULL;
if( di && jl >= r + 1 )
{
if( temp < statL[ii][jl].first )
{
temp = statL[ii][jl].first;
temp_num = statL[ii][jl].second;
}
else if( temp == statL[ii][jl].first )
temp_num += statL[ii][jl].second;
}
if( jr <= m - r )
{
if( temp < statR[ii][jr].first )
{
temp = statR[ii][jr].first;
temp_num = statR[ii][jr].second;
}
else if( temp == statR[ii][jr].first )
temp_num += statR[ii][jr].second;
}
}
if( !temp_num )
continue;
if(ans < temp + circle[i][j] )
{
ans = temp + circle[i][j];
num = temp_num;
}
else if( ans == temp + circle[i][j] )
num += temp_num;
}
}
cout << ans << " " << num << endl;
}
int main()
{
while( cin >> n >> m >> r )
{
for(int i = 1; i <= n; ++i)
{
for(int j = 1; j <= m; ++j)
{
cin >> cell[i][j];
accum[i][j] = accum[i][j-1] + cell[i][j];
}
}
Init();
Work();
}
return 0;
}