题目：http://hihocoder.com/problemset/problem/1037
- 一个迷宫有n层，第 i 层有 i 个房间
- 从第i层的第i个房间(i, i)可以走到第i+1层的第i个房间或第i+1房间(i+1, i)/(i+1, i+1)
- 每个房间走出去后都不能回头
样例输入
5 //迷宫的层数 2 6 4 1 2 8 4 0 9 6 6 5 5 3 6
结果： 28
思路
动态规划
maze[i, j] = max(maze[i-1, j], maze[i-1, j-1])
源码
- 迭代方式，从下而上
1 #include <iostream> 2 using namespace std; 3 4 static int maze[100][100];//存储到达每一个房间收集到的最多券 5 static int data[100][100];//存储每一个房间的券 6 7 int main() 8 { 9 int maxLevel, i = 0, max = 0; 10 cin >> maxLevel; 11 12 for (i = 0; i < maxLevel; i++) 13 for (int j = 0; j <= i; j++) 14 cin >> data[i][j]; 15 //由低层到高层计算 16 maze[0][0] = data[0][0]; 17 for (int i = 1; i < maxLevel; i++) 18 { 19 for (int j = 0; j <= i; j++) 20 { 21 if (j > 0 && maze[i - 1][j] < maze[i - 1][j - 1]) 22 maze[i][j] = maze[i - 1][j - 1]; 23 else 24 maze[i][j] = maze[i - 1][j]; 25 maze[i][j] += data[i][j]; 26 } 27 } 28 //找出最后一层中券最多的房间 29 for (i = 0; i < maxLevel; i++) 30 { 31 int tmp = maxLevel - 1; 32 if (max < maze[tmp][i]) 33 max = maze[tmp][i]; 34 } 35 cout << max << endl; 36 return 0; 37 }
- 递归方式，从上而下
1 #include <iostream> 2 using namespace std; 3 4 int maze[100][100]; 5 int data[100][100]; 6 void dp(int level) 7 { 8 if (level < 0) 9 return; 10 dp(level - 1); 11 for (int room = 0; room <= level; room++) 12 { 13 if (level > 0) 14 { 15 if (room > 0) 16 maze[level][room] = maze[level - 1][room] > maze[level - 1][room - 1] ? maze[level - 1][room] : maze[level - 1][room - 1]; 17 else 18 maze[level][room] = maze[level - 1][room]; 19 } 20 else 21 maze[level][room] = 0; 22 maze[level][room] += data[level][room]; 23 } 24 } 25 int main() 26 { 27 int maxLevel, i = 0, max = 0; 28 cin >> maxLevel; 29 30 for (i = 0; i < maxLevel; i++) 31 for (int j = 0; j <= i; j++) 32 cin >> data[i][j]; 33 dp(maxLevel - 1); 34 for (i = 0; i < maxLevel; i++) 35 { 36 int tmp = maxLevel - 1; 37 if (max < maze[tmp][i]) 38 max = maze[tmp][i]; 39 } 40 cout << max << endl; 41 return 0; 42 }