hdu 5569(二维dp，水题)

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题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=5569

解题思路：dp[i][j]表示到达(i,j)位置时的最小值。由于目标式的特点，我们一次需要走两格，自然而然转移方程也是两次走：

dp[i][j]->dp[i+2][j]，dp[i][j]->dp[i][j+2]，dp[i][j]->dp[i+1][j+1]。属于简单的dp问题

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 1005;const int inf = 0x3f3f3f3f;int n,m,mat[maxn][maxn];int dp[maxn][maxn];int main(){while(scanf("%d%d",&n,&m)!=EOF){memset(dp,inf,sizeof(dp));for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++)scanf("%d",&mat[i][j]);dp[1][2] = mat[1][1] * mat[1][2];dp[2][1] = mat[1][1] * mat[2][1];for(int i = 1; i <= n; i++)for(int j = 1; j <= m; j++){if(dp[i][j] == inf) continue;if(i + 2 <= n)dp[i+2][j] = min(dp[i+2][j],dp[i][j] + mat[i+1][j] * mat[i+2][j]);if(j + 2 <= m)dp[i][j+2] = min(dp[i][j+2],dp[i][j] + mat[i][j+1] * mat[i][j+2]);if(i + 1 <= n && j + 1 <= m){dp[i+1][j+1] = min(dp[i+1][j+1],dp[i][j] + mat[i][j+1] * mat[i+1][j+1]);dp[i+1][j+1] = min(dp[i+1][j+1],dp[i][j] + mat[i+1][j] * mat[i+1][j+1]);}}printf("%d\n",dp[n][m]);}return 0;}

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