HDU 5569 matrix

题目传送门

中文翻译:

解题思路:

f[i][j]表示到第i行第j列时,可获得的最小收益

AC代码:

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 
 5 using namespace std;
 6 
 7 long long n,m,a[1001][1001],f[1001][1001]; 
 8 
 9 int main() {
10     while(scanf("%d%d",&n,&m) != EOF) {
11         memset(a,0,sizeof(a));
12         memset(f,0x3f3f3f,sizeof(f));
13         for(int i = 1;i <= n; i++)
14             for(int j = 1;j <= m; j++) 
15                 scanf("%d",&a[i][j]);
16         f[1][1] = a[1][1];
17         for(int i = 2;i <= m; i++) {//处理第一行 
18             if(i % 2 == 0) f[1][i] = f[1][i-1] + a[1][i] * a[1][i-1];
19             else f[1][i] = f[1][i-1];
20             if(i == 2) f[1][2] -= f[1][1];
21         }
22         for(int i = 2;i <= n; i++) {//处理第一列 
23             if(i % 2 == 0) f[i][1] = f[i-1][1] + a[i][1] * a[i-1][1];
24             else f[i][1] = f[i-1][1];
25             if(i == 2) f[2][1] -= f[1][1];
26         }
27         for(int i = 2;i <= n; i++)
28             for(int j = 2;j <= m; j++) {
29                 if((j + i) % 2 == 0)//走了奇数步 
30                     f[i][j] = min(f[i-1][j],f[i][j-1]);
31                 else //走了偶数步 
32                     f[i][j] = min(f[i-1][j] + a[i-1][j] * a[i][j],f[i][j-1] + a[i][j-1] * a[i][j]);
33             }
34         printf("%d\n",f[n][m]);
35     }
36     return 0;
37 }