Contents
Problem
找出炸彈可以殺掉最多老鼠的位置和數量。
Solution
記錄每一格可以殺的老鼠數量，也就是將可以炸到該群老鼠的位置，都加上當前老鼠數。
另外一種方法則是把老鼠確切的位置存入它的數量，以移動炸彈大小的方格去找出最大和。(比較慢)
Code
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using namespace std;
inline int get_num()
{
int n = 0;
char c;
while ((c = getchar()) != '\n' && c != ' ' && c != EOF)
n = n * 10 + c - '0';
return n;
}
int main()
{
static int grid[N][N] = {};
int T;
scanf("%d", &T);
while (T--)
{
int n, d;
int lx = N, ly = N, rx = 0, ry = 0; //邊界
scanf("%d%d ", &d, &n);
int mx = 0, my = 0, ans = 0;
for (int i = 0; i < n; ++i)
{
int x, y, size;
x = get_num();
y = get_num();
size = get_num();
lx = min(lx, x - d);
ly = min(ly, y - d);
rx = max(rx, x + d);
ry = max(ry, y + d);
int xa = max(0, x - d), ya = max(0, y - d), xb = min(1024, x + d), yb = min(1024, y + d);
for (int i = xa; i <= xb; ++i)
for (int j = ya; j <= yb; ++j)
grid[i][j] += size; //當炸彈放在 (i, j) 時可消滅的老鼠數量
}
lx = max(0, lx);
ly = max(0, ly);
rx = min(N - 1, rx);
ry = min(N - 1, ry);
for (int x = lx; x <= rx; ++x)
{
for (int y = ly; y <= ry; ++y)
if (ans < grid[x][y])
{
ans = grid[x][y];
mx = x;
my = y;
}
}
printf("%d %d %d\n", mx, my, ans);
for (int i = lx; i <= rx; ++i)
memset(grid[i], 0, sizeof grid[i]);
}
return 0;
}
方法二1
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using namespace std;
int main()
{
static int grid[N][N] = {}, kill[N][N] = {};
int T;
scanf("%d", &T);
while (T--)
{
int n, d, bottom = 0;
int lx = N, ly = N, rx = 0, ry = 0; //邊界
scanf("%d%d", &d, &n);
for (int i = 0; i < n; ++i)
{
int x, y, size;
scanf("%d%d%d", &x, &y, &size);
lx = min(lx, x);
ly = min(ly, y);
rx = max(rx, x);
ry = max(ry, y);
bottom = max(bottom, x); //init 用
grid[x][y] = size;
}
lx = max(0, lx - d);
ly = max(0, ly - d);
rx = min(N - 1, rx + d);
ry = min(N - 1, ry + d);
for (int x = lx; x <= rx; ++x)
{
//初始 window (ly-d~ly+d) 內的值
int left = max(0, ly - d);
for (int i = left; i <= ly + d && i < N; ++i)
kill[x][ly] += grid[x][i];
//移動 window 並增減頭尾的數字
for (int i = ly + 1; i <= ry; ++i)
kill[x][i] = kill[x][i - 1] - ((i - d - 1 >= 0) ? grid[x][i - d - 1] : 0) + ((i + d < N) ? grid[x][i + d] : 0);
}
int ans = 0, mx = 0, my = 0;
for (int x = lx; x <= rx; ++x)
for (int y = ly; y <= ry; ++y)
{
int sum = 0;
//把炸彈範圍內列的相加
for (int i = max(0, x - d); i <= x + d; ++i)
sum += kill[i][y];
if (sum > ans)
{
ans = sum;
mx = x;
my = y;
}
}
printf("%d %d %d\n", mx, my, ans);
for (int i = lx; i <= rx; ++i)
{
memset(grid[i], 0, sizeof grid[i]);
memset(kill[i], 0, sizeof kill[i]);
}
}
return 0;
}