HDU5569(dp)

matrix

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 598 Accepted Submission(s): 345



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 is a1∗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)

Recommend

hujie

分析：和hihocoder第五周的的题类似，不过这个多了2种情况，对边界特判即可。dp[i][j]表示到达这一点的最小值，并且i+j必须是奇数。

代码：

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
using namespace std;
long long dp[1005][1005];
int a[1005][1005];
int n,m;
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
scanf("%d",&a[i][j]);
}
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if((i+j)%2==0) continue;
if(i==1)
{
dp[i][j]=dp[1][j-2]+a[1][j-1]*a[1][j];
continue;
}
if(j==1)
{
dp[i][j]=a[i-1][1]*a[i][1]+dp[i-2][1];
continue;
}
if(j!=1&&i!=1)
{
dp[i][j]=min((long long)dp[i-1][j-1]+a[i-1][j]*a[i][j],(long long)dp[i-1][j-1]+a[i][j-1]*a[i][j]);
if(i-2>=1)
{
dp[i][j]=min(dp[i][j],dp[i-2][j]+(long long)a[i-1][j]*a[i][j]);
}
if(j-2>=1)
{
dp[i][j]=min(dp[i][j],dp[i][j-2]+(long long)a[i][j-1]*a[i][j]);
}
}
}
}
printf("%I64d\n",dp
[m]);
}
}