HDU 5569 matrix

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 isa1∗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

给一个n*m矩阵(n+m为奇数) 从(1,1)走到(n,m) 每次只能走右边和下边 
走过的路会得到值 a1,a2,a3,...an
求 a1*a2 + a3*a4 + ... ak-1*ak 最小值 


我用的dp+记忆化搜索
在走第奇数步的时候 dp[i][j] = min(dp[i][j+1]+mp[i][j]*mp[i][j+1], dp[i+1][j]+mp[i][j]*mp[i+1][j])
在走第偶数步的时候 dp[i][j] = min(dp[i][j+1], dp[i+1][j]) 

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <cmath>#include <stack>#include <string>#include <map>#include <set>#define pi acos(-1)#define LL long long#define ULL unsigned long long#define inf 0x3f3f3f3f#define INF 1e18#define lson l,mid,rt<<1#define rson mid+1,r,rt<<1|1using namespace std;typedef pair<int, int> P;const int maxn = 1e3 + 5;const int mod = 1e8;int mp[maxn][maxn];int dp[maxn][maxn];int dir[2][2] = {{1,0}, {0,1}};int n, m;int dfs(int x, int y, int cnt){    if (dp[x][y]) return dp[x][y];    if (x == n && y == m){        return 0;    }    int tmp, mni = inf;    for (int i = 0; i < 2; i++){        int nx = x + dir[i][0];        int ny = y + dir[i][1];        if (nx > n || ny > m) continue;        if (cnt % 2){            tmp = dfs(nx, ny, cnt+1);            tmp = tmp + mp[x][y]*mp[nx][ny];            mni = min(mni, tmp);        }        else {            tmp = dfs(nx, ny, cnt+1);            mni = min(mni, tmp);        }    }    dp[x][y] = mni;    return dp[x][y];}int main(void){//freopen("C:\\Users\\wave\\Desktop\\NULL.exe\\NULL\\in.txt","r", stdin);    int i, j, ans;    while (cin >> n >> m){        memset(dp, 0, sizeof(dp));        for (i = 1; i <= n; i++)            for (j = 1; j <= m; j++)                scanf("%d", &mp[i][j]);        ans = dfs(1, 1, 1);        printf("%d\n", ans);    }return 0;}