递推DP HDOJ 5092 Seam Carving

题目传送门

 1 /*
 2     题意：从上到下，找最短路径，并输出路径
 3     DP：类似数塔问题，上一行的三个方向更新dp，路径输出是关键
 4 */
 5 #include <cstdio>
 6 #include <algorithm>
 7 #include <iostream>
 8 #include <cstring>
 9 #include <cmath>
10 #include <string>
11 #include <vector>
12 #include <queue>
13 #include <map>
14 #include <set>
15 #include <ctime>
16 #include <cstdlib>
17 using namespace std;
18
19 const int MAXN = 1e2 + 10;
20 const int INF = 0x3f3f3f3f;
21 int a[MAXN][MAXN];
22 int dp[MAXN][MAXN];
23 int pre[MAXN][MAXN];
24
25 void print(int x, int y)
26 {
27     if (x == 1)
28     {
29         printf ("%d", y);    return ;
30     }
31     print (x - 1, pre[x][y]);
32     printf (" %d", y);
33 }
34
35 int main(void)        //HDOJ 5092 Seam Carving
36 {
37     //freopen ("C.in", "r", stdin);
38
39     int n, m, t, cas = 0;
40     scanf ("%d", &t);
41     while (t--)
42     {
43         scanf ("%d%d", &n, &m);
44         for (int i=1; i<=n; ++i)
45             for (int j=1; j<=m; ++j)    scanf ("%d", &a[i][j]);
46         memset (pre, 0, sizeof (pre));
47
48         for (int i=2; i<=n; ++i)
49             for (int j=1; j<=m; ++j)    dp[i][j] = INF;
50         for (int i=1; i<=m; ++i)    dp[1][i] = a[1][i];
51
52         for (int i=2; i<=n; ++i)
53         {
54             for (int j=m; j>=1; --j)
55             {
56                 for (int k=1; k>=-1; --k)
57                 {
58                     if (j + k < 1 || j + k > m)    continue;
59                     if (dp[i][j] > dp[i-1][j+k] + a[i][j])
60                     {
61                         dp[i][j] = dp[i-1][j+k] + a[i][j];
62                         pre[i][j] = j + k;
63                     }
64                 }
65             }
66         }
67
68         int k = m;    int mn = dp[n][m];
69         for (int i=m-1; i>=1; --i)
70         {
71             if (mn > dp[n][i])    {mn = dp[n][i];    k = i;}
72         }
73
74         printf ("Case %d\n", ++cas);
75         print (n, k);
76         puts ("");
77     }
78
79     return 0;
80 }
81
82 /*
83 Case 1
84 2 1 1 2
85 Case 2
86 3 2 1 1 2 1
87 */