题目链接：http://poj.org/problem?id=1163
题目思路：
从三角形的底部开始考虑
1 #include <iostream> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <cstdio> 6 using namespace std; 7 const int MAX = 100; 8 int a[MAX][MAX], dp[MAX][MAX]; 9 int main(void) { 10 //freopen("1163.in", "r", stdin); 11 int n,i,j; scanf("%d",&n); 12 for (i = 0; i < n; ++i) { 13 for (j = 0; j < i+1; ++j) { 14 scanf("%d", &a[i][j]); 15 } 16 } 17 memset(dp, 0, sizeof(dp)); 18 for (i = 0; i < n; ++i) dp[n-1][i] = a[n-1][i]; 19 for (i = n - 2; i >= 0; --i) { 20 for (j = 0; j < n - 1; ++j) { 21 dp[i][j] = max(dp[i+1][j], dp[i+1][j+1]) + a[i][j]; 22 } 23 } 24 printf("%d\n", dp[0][0]); 25 return 0; 26 }
还是从最简单的开始做起吧
另外一种写法：
1 #include <iostream> 2 #include <cstdlib> 3 #include <cstring> 4 #include <algorithm> 5 #include <cstdio> 6 using namespace std; 7 const int MAX = 100; 8 int a[MAX][MAX], dp[MAX][MAX], n; 9 int f(int i, int j) { 10 if (dp[i][j] >= 0) return dp[i][j]; 11 else { 12 if (i == n-1) dp[i][j] = a[i][j]; 13 else dp[i][j] = a[i][j] + max(dp[i+1][j], dp[i+1][j+1]); 14 return dp[i][j]; 15 } 16 } 17 int main(void) { 18 int i,j; scanf("%d",&n); 19 for (i = 0; i < n; ++i) { 20 for (j = 0; j < i+1; ++j) { 21 scanf("%d", &a[i][j]); 22 } 23 } 24 memset(dp, -1, sizeof(dp)); 25 for (i = 0; i < n; ++i) dp[n-1][i] = a[n-1][i]; 26 for (i = n-1; i >= 0; --i) 27 for (j = 0; j < i+1; ++j) 28 f(i, j); 29 printf("%d\n", dp[0][0]); 30 return 0; 31 }
这就是所谓的记忆化搜索
