hdu 5569 matrix(简单dp)

Problem Description

Given a matrix with n rows and m columns ( n+m is an odd number ), at first , you begin with the number at top-left corner (,) and you want to go to the number at bottom-right corner (n,m). And you must go right or go down every steps. Let the numbers you go through become an array a1,a2,...,a2k. The cost isa1∗a2+a3∗a4+...+a2k−∗a2k. What is the minimum of the cost?
 

Input

Several test cases(about )

For each cases, first come  integers, n,m(≤n≤,≤m≤)

N+m is an odd number.

Then follows n lines with m numbers ai,j(≤ai≤)

 

Output

For each cases, please output an integer in a line as the answer.

 

Sample Input

 

Sample Output

 

Source

BestCoder Round #63 (div.2)

令dp[i][j]表示当前走到第i,j个位置的最小贡献，初始化做好了，然后根据i+j是奇数偶数的情况分别计算dp即可，最后要用long long。

 #pragma comment(linker, "/STACK:1024000000,1024000000")
 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<bitset>
 #include<map>
 #include<vector>
 #include<stdlib.h>
 #include <stack>
 using namespace std;
 #define PI acos(-1.0)
 #define max(a,b) (a) > (b) ? (a) : (b)
 #define min(a,b) (a) < (b) ? (a) : (b)
 #define ll long long
 #define eps 1e-10
 #define MOD 1000000007
 #define N 1006
 #define inf 1<<29
 ll n,m;
 ll mp[N][N];
 ll dp[N][N];
 int main()
 {
    while(scanf("%I64d%I64d",&n,&m)==){
          memset(dp,,sizeof(dp));
       for(ll i=;i<=n;i++){
          for(ll j=;j<=m;j++){
             scanf("%I64d",&mp[i][j]);
          }
       }
       for(ll i=;i<=n+;i++){
          dp[i][]=inf;
          mp[i][]=inf;
       }
       for(ll i=;i<=m+;i++){
          dp[][i]=inf;
          mp[][i]=inf;
       }

       dp[][]=mp[][];
       dp[][]=mp[][]*mp[][];
       dp[][]=mp[][]*mp[][];
       for(ll i=;i<=n;i++){
          for(ll j=;j<=m;j++){
             if(i== && j==) continue;
             if(i== && j==) continue;
             if(i== && j==) continue;
             if((i+j)%==){
                dp[i][j]=min(dp[i-][j],dp[i][j-]);
                //printf("i , j %d %d %d\n",i,j,dp[i][j]);
             }
             else{
                dp[i][j]=min(dp[i-][j]+mp[i-][j]*mp[i][j],dp[i][j-]+mp[i][j-]*mp[i][j]);
                //printf("i , j %d %d %d\n",i,j,dp[i][j]);
             }

          }
       }

       printf("%I64d\n",dp[n][m]);
    }
     return ;
 }