HDOJ 5569 matrix (DP)

matrix

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

Total Submission(s): 340    Accepted Submission(s): 208


[align=left]Problem Description[/align]
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?
 

[align=left]Input[/align]
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)

 

[align=left]Output[/align]
For each cases, please output an integer in a line as the answer.
 

[align=left]Sample Input[/align]

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

 

[align=left]Sample Output[/align]

4
8

题意：应该都能看懂，就不再翻译了

思路：刚开始看，没想到dp，写完才发现，因为dp弱，所以推了好久，因为只能向右或者向下，所以下一步的状态是由上一步来的，dp数组记录了到达那个点的ans最小值，然后一步一步推，相当于记忆化，在起点的时候该点dp值为0，周围边界直接处理为INF。
在奇数步的时候：转移方程为：dp[i][j]=min(dp[i-1][j],dp[i][j-1]);
在偶数步的时候：转移方程为：dp[i][j]=min(dp[i-1][j]+map[i-1][j]*map[i][j],dp[i][j-1]+map[i][j-1]*map[i][j]);

ac代码：

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
#include<algorithm>
#define MAXN 10000001
#define fab(a) (a)>0?(a):(-a)
#define INF 0xfffffff
#define LL long long
#define mem(x) memset(x,0,sizeof(x))
using namespace std;
int map[1010][1010];
LL dp[1010][1010];
int n,m;
int main()
{
int i,j;
while(~scanf("%d%d",&n,&m))
{
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
scanf("%d",&map[i][j]);
}
for(i=0;i<=n+1;i++)
{
for(j=0;j<=m+1;j++)
dp[i][j]=INF;
}
dp[1][1]=0;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(i==1&&j==1)
continue;
if((i+j)%2==0)
dp[i][j]=min(dp[i-1][j],dp[i][j-1]);
else
dp[i][j]=min(dp[i-1][j]+map[i-1][j]*map[i][j],dp[i][j-1]+map[i][j-1]*map[i][j]);

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