TYVJ1124 花店橱窗

　　　http://www.tyvj.cn/Problem_Show.aspx?id=1124

　　设状态dp[i][j] 为第i个花放入第j个瓶子里所能得到的最大价值  且  (i<=j<=i+m-n)

　　dp[i][j] = max{dp[i-1][k]  |  i-1<=k<j}  + a[i][j] （a[i][j]为第i个花放入第j个瓶的价值）

　　每次转移的时候记录下路径，输出的时候dfs回去就ok了~

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <algorithm>
 5 #include <vector>
 6 #include <cmath>
 7 #include <string>
 8 #include <map>
 9 #include <iostream>
10 using namespace std;
11 typedef long long LL;
12 const int maxn = 105;
13 const int maxv = 1005;
14 const int mod = 1e9+7;
15 const int INF = 0x3f3f3f3f;
16 int dp[maxn][maxn],a[maxn][maxn];
17 int path[maxn][maxn];
18 int n,m;
19 void dfs(int i,int j)
20 {
21     if(i==1)
22     {
23         printf("%d ",j);
24         return;
25     }
26     dfs(i-1,path[i][j]);
27     printf("%d",j);
28     if(i!=n)printf(" ");
29 }
30 int main()
31 {
32 //    freopen("in.txt","r",stdin);
33     while(cin>>n>>m)
34     {
35         for(int i = 1;i<=n;++i)
36             for(int j = 1;j<=m;++j)
37                 scanf("%d",&a[i][j]);
38         memset(dp,0,sizeof(dp));
39         memset(path,-1,sizeof(path));
40         int x = m-n;
41         for(int i = 1;i<=n;++i)
42         {
43             int up = i+x;
44             for(int j = i;j<=up;++j)
45             {
46                 dp[i][j] = dp[i-1][i-1]+a[i][j];
47                 path[i][j] = i-1;
48                 for(int k = i;k<j;++k)
49                 {
50                     if(dp[i][j]<dp[i-1][k]+a[i][j])
51                     {
52                         dp[i][j] = dp[i-1][k]+a[i][j];
53                         path[i][j] = k;
54                     }
55                 }
56             }
57         }
58         int ans = -INF,ans_x;
59         for(int i = n;i<=m;++i)if(ans<dp[n][i])
60         {
61             ans = dp[n][i];
62             ans_x = i;
63         }
64         cout<<ans<<endl;
65         dfs(n,ans_x);
66         puts("");
67     }
68     return 0;
69 }