matrix（dp）

matrix

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 91    Accepted Submission(s): 62

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)

 

题解：dp题，发现自己对于dp太弱了。。。

题意是一个n*m的矩阵，从（1，1）走到（n,m）的a1*a2+a3*a4+a5*a6。。。。其中a1,a2。。。是矩阵的当前值；

由于只能向右向下，从（1，1）开始，那么(x+1,y)或（x,y+1）由于是从1+1=2偶数开始，所以到x+y+1是奇数的时候再进行运算，

所以每次往前推两步即可，初始化dp为INF，因为求最小值，dp[1][1]=0；状态转移方程：

if((i+j)&1){
    dp[i][j]=min(dp[i-1][j]+num[i-1][j]*num[i][j],dp[i][j-1]+num[i][j-1]*num[i][j]);
   }
   else dp[i][j]=min(dp[i-1][j],dp[i][j-1]);

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#define mem(x,y) memset(x,y,sizeof(x))
using namespace std;
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const int MAXN=1010;
int dp[MAXN][MAXN],num[MAXN][MAXN];
int main(){
	int n,m;
	while(~scanf("%d%d",&n,&m)){
		mem(dp,INF);
		for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++){
			scanf("%d",&num[i][j]);
			if(i==1&&j==1){
				dp[i][j]=0;continue;
			}
			if((i+j)&1){
				dp[i][j]=min(dp[i-1][j]+num[i-1][j]*num[i][j],dp[i][j-1]+num[i][j-1]*num[i][j]);
			}
			else dp[i][j]=min(dp[i-1][j],dp[i][j-1]);
		}
		printf("%d\n",dp[n][m]);
	}
	return 0;
}