각 셀에 포인트가 있는 행렬이 있으며, 두 번의 탐색을 사용하여 해당 그리드에서 최대 포인트를 얻는 방법입니다.
충족해야 할 몇 가지 조건이 있습니다 -
- 첫 번째 순회는 그리드의 왼쪽 상단 셀에서 시작하여 왼쪽 하단 모서리로 이동해야 합니다. 그리고 두 번째 순회에서 오른쪽 상단 모서리에서 시작하여 오른쪽 하단 모서리로
- 한 셀에서 맨 아래, 현재 셀의 왼쪽 아래, 현재 셀의 오른쪽 아래로만 이동할 수 있습니다.
- 한 순회가 이미 셀에서 일부 포인트를 얻은 경우 다음 순회에서는 해당 셀에서 포인트를 얻지 못합니다.
입력 및 출력
Input: A grid with points. 3 6 8 2 5 2 4 3 1 1 20 10 1 1 20 10 1 1 20 10 Output: Maximum points collected by two traversals is 73. From the first traversal, it gains: 3 + 2 + 20 + 1 + 1 = 27 From the second traversal, it gains: 2 + 4 + 10 + 20 + 10 = 46
알고리즘
findMaxVal(mTable, x, y1, y2)
입력 - 3D 배열을 암기 테이블로, x 값과 y1, y2.
출력 - 최대값.
Begin if x, y1 and y2 is not valid, then return - ∞ if both traversal is complete, then if y1 = y2, then return grid[x, y1] else return grid[x, y1] + grid[x, y2] if both traversal are at last row, then return - ∞ if subProblem is solved, then return mTable[x, y1, y2] set res := - ∞ if y1 = y2, then temp := grid[x, y1] else temp := grid[x, y1] + grid[x, y2] res := max of res and (temp + findMaxVal(mTable, x+1, y1, y2-1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1, y2+1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1, y2)) res := max of res and (temp + findMaxVal(mTable, x+1, y1-1, y2)) res := max of res and (temp + findMaxVal(mTable, x+1, y1-1, y2-1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1-1, y2+1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1+1, y2)) res := max of res and (temp + findMaxVal(mTable, x+1, y1+1, y2-1)) res := max of res and (temp + findMaxVal(mTable, x+1, y1+1, y2+1)) return true if mTable[x, y1, y2] = res End인 경우 true를 반환합니다.
예
#include<iostream>
#define ROW 5
#define COL 4
using namespace std;
int grid[ROW][COL] = {
{3, 6, 8, 2},
{5, 2, 4, 3},
{1, 1, 20, 10},
{1, 1, 20, 10},
{1, 1, 20, 10},
};
bool isValidInput(int x, int y1, int y2) {
return (x >= 0 && x < ROW && y1 >=0 && y1 < COL && y2 >=0 && y2 < COL);
}
int max(int a, int b) {
return (a>b)?a:b;
}
int findMaxVal(int mTable[ROW][COL][COL], int x, int y1, int y2) {
if (!isValidInput(x, y1, y2)) //when in invalid cell, return -ve infinity
return INT_MIN;
if (x == ROW-1 && y1 == 0 && y2 == COL-1) //when both traversal is complete
return (y1 == y2)? grid[x][y1]: grid[x][y1] + grid[x][y2];
if (x == ROW-1) //both traversal are at last row but not completed
return INT_MIN;
if (mTable[x][y1][y2] != -1) //when subproblem is solved
return mTable[x][y1][y2];
int answer = INT_MIN; //initially the answer is -ve infinity
int temp = (y1 == y2)? grid[x][y1]: grid[x][y1] + grid[x][y2]; //store gain of the current room
//find answer for all possible value and use maximum of them
answer = max(answer, temp + findMaxVal(mTable, x+1, y1, y2-1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1, y2+1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1, y2));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1-1, y2));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1-1, y2-1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1-1, y2+1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1+1, y2));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1+1, y2-1));
answer = max(answer, temp + findMaxVal(mTable, x+1, y1+1, y2+1));
return (mTable[x][y1][y2] = answer); //store the answer in the mTable and return.
}
int findMaxCollection() {
// Create a memoization table and set all values as -1
int mTable[ROW][COL][COL];
for(int i = 0; i<ROW; i++)
for(int j = 0; j<COL; j++)
for(int k = 0; k<COL; k++)
mTable[i][j][k] = -1;
return findMaxVal(mTable, 0, 0, COL-1);
}
int main() {
cout << "Maximum collection is " << findMaxCollection();
return 0;
} 출력
Maximum collection is 73