HDU-5569 matrix（DP）

matrix

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

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 31 2 32 2 12 32 2 11 2 4

 

Sample Output

48

很快就想到了DP，题真不是白做的

对于任意i,j，且i+j为奇数时，dp[i][j]只能从dp[i-2][j],dp[i-1][j-1],dp[i][j-2]转移过来。

其中:

①a[i-2][j]只能经过a[i-1][j]到达a[i][j]，状态转移方程为：dp[i][j]=min(dp[i][j],dp[i-2][j],a[i-1][j]*a[i][j]);

②a[i][j-2]只能经过a[i][j-1]到达a[i][j]，状态转移方程为：dp[i][j]=min(dp[i][j],dp[i][j-2],a[i][j-1]*a[i][j]);

③a[i][j]可分别经过a[i][j-1]和a[i-1][j]到达a[i][j]，状态转移方程为：

dp[i][j]=min(dp[i][j],dp[i-1][j-1]+a[i][j]*min(a[i-1][j],a[i][j-1]));

合并后可得：

dp[i][j]=min(dp[i][j],min(min(dp[i-1][j-1],dp[i-2][j])+a[i][j]*a[i-1][j],min(dp[i-1][j-1],dp[i][j-2])+a[i][j]*a[i][j-1]));

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;int a[1005][1005],dp[1005][1005],n,m;int main() {    int i,j;    while(scanf("%d%d",&n,&m)==2) {        memset(dp,0x3f,sizeof(dp));       dp[0][1]=dp[1][0]=0;        for(i=1;i<=n;++i)            for(j=1;j<=m;++j)                scanf("%d",&a[i][j]);        for(i=2;i<=n;++i)//先处理第1列，防止越界            dp[i][1]=dp[i-2][1]+a[i][1]*a[i-1][1];        for(j=2;j<=m;++j)//先处理第1行，防止越界            dp[1][j]=dp[1][j-2]+a[1][j]*a[1][j-1];        //没想到用“我为人人”型转移状态，这样就可以直接处理，不用担心越界了        for(i=2;i<=n;++i)            for(j=2;j<=m;++j)                if((i+j)&1)                    dp[i][j]=min(dp[i][j],min(min(dp[i-1][j-1],dp[i-2][j])+a[i][j]*a[i-1][j],min(dp[i-1][j-1],dp[i][j-2])+a[i][j]*a[i][j-1]));        printf("%d\n",dp[n][m]);    }    return 0;}