AtCoder Beginner Contest 147 E. Balanced Path

思路：

dp，使用了bitset优化。

实现：

 1 #include 
 2 using namespace std;
 3 const int N = 81;
 4 const int INF = 0x3f3f3f3f;
 5 int a[N][N], b[N][N];
 6 bitset4> dp[N][N];
 7 
 8 int main()
 9 {
10     int h, w;
11     while (cin >> h >> w)
12     {
13         for (int i = 0; i < h; i++)
14         {
15             for (int j = 0; j < w; j++)
16             {
17                 cin >> a[i][j];
18             }
19         }
20         for (int i = 0; i < h; i++)
21         {
22             for (int j = 0; j < w; j++)
23             {
24                 cin >> b[i][j];
25             }
26         }
27         for (int i = 0; i < h; i++)
28         {
29             for (int j = 0; j < w; j++)
30             {
31                 dp[i][j].reset();
32             }
33         }
34         dp[0][0][a[0][0] - b[0][0] + N * N * 2] = 1;
35         dp[0][0][b[0][0] - a[0][0] + N * N * 2] = 1;
36         for (int i = 0; i < h; i++)
37         {
38             for (int j = 0; j < w; j++)
39             {
40                 int tmp = abs(a[i][j] - b[i][j]);
41                 if (j)
42                 {
43                     dp[i][j] |= dp[i][j - 1] << tmp;
44                     dp[i][j] |= dp[i][j - 1] >> tmp;
45                 }
46                 if (i)
47                 {
48                     dp[i][j] |= dp[i - 1][j] << tmp;
49                     dp[i][j] |= dp[i - 1][j] >> tmp;
50                 }
51             }
52         }
53         int res = INF;
54         for (int i = 0; i < N * N * 4; i++)
55         {
56             if (dp[h - 1][w - 1][i]) res = min(res, abs(i - N * N * 2));
57         }
58         cout << res << endl;
59     }
60     return 0;
61 }