hdoj--5569--matrix(动态规划)

matrix

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

Total Submission(s): 509 Accepted Submission(s): 297

[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

[align=left]Source[/align]
BestCoder Round #63 (div.2)

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用dp数组记录下到达每一个点最小的路径

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
long long dp[1010][1010],num[1010][1010];
int m,n;
int main()
{
while(scanf("%d%d",&n,&m)!=EOF)
{
memset(num,0,sizeof(num));
memset(dp,0,sizeof(dp));
for(int i=1;i<=n;i++)
for(int j=1;j<=m;j++)
scanf("%lld",&num[i][j]);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(i==1&&j==1)
dp[i][j]=0;
else if((i+j)&1)
{
if(i==1)
dp[i][j]=dp[i][j-1]+num[i][j-1]*num[i][j];
else if(j==1)
dp[i][j]=dp[i-1][j]+num[i-1][j]*num[i][j];
else
dp[i][j]=min(dp[i][j-1]+num[i][j-1]*num[i][j],dp[i-1][j]+num[i-1][j]*num[i][j]);
}
else
{
if(i==1)
dp[i][j]=dp[i][j-1];
else if(j==1)
dp[i][j]=dp[i-1][j];
else dp[i][j]=min(dp[i-1][j],dp[i][j-1]);
}
}
}
printf("%lld\n",dp
[m]);
}
return 0;
}