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 (1,1) 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−1∗a2k. What is the minimum of the cost?
 

Input

Several test cases(about 5)

For each cases, first come 2 integers, n,m(1≤n≤1000,1≤m≤1000)

N+m is an odd number.

Then follows n lines with m numbers ai,j(1≤ai≤100)

 

Output

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

 

Sample Input

2 3
1 2 3
2 2 1
2 3
2 2 1
1 2 4

 

Sample Output

4 
8

 

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)==2){
         memset(dp,0,sizeof(dp));
      for(ll i=1;i<=n;i++){
         for(ll j=1;j<=m;j++){
            scanf("%I64d",&mp[i][j]);
         }
      }
      for(ll i=0;i<=n+1;i++){
         dp[i][0]=inf;
         mp[i][0]=inf;
      }
      for(ll i=0;i<=m+1;i++){
         dp[0][i]=inf;
         mp[0][i]=inf;
      }

      dp[1][1]=mp[1][1];
      dp[1][2]=mp[1][1]*mp[1][2];
      dp[2][1]=mp[1][1]*mp[2][1];
      for(ll i=1;i<=n;i++){
         for(ll j=1;j<=m;j++){
            if(i==1 && j==1) continue;
            if(i==1 && j==2) continue;
            if(i==2 && j==1) continue;
            if((i+j)%2==0){
               dp[i][j]=min(dp[i-1][j],dp[i][j-1]);
               //printf("i , j %d %d %d
",i,j,dp[i][j]);
            }
            else{
               dp[i][j]=min(dp[i-1][j]+mp[i-1][j]*mp[i][j],dp[i][j-1]+mp[i][j-1]*mp[i][j]);
               //printf("i , j %d %d %d
",i,j,dp[i][j]);
            }

         }
      }

      printf("%I64d
",dp[n][m]);
   }
    return 0;
}