【动态规划】HDU 5492 Find a path (2015 ACM/ICPC Asia Regional Hefei Online)

 1 //
 2 //by coolxxx
 3 //#include<bits/stdc++.h>
 4 #include<iostream>
 5 #include<algorithm>
 6 #include<string>
 7 #include<iomanip>
 8 #include<map>
 9 #include<stack>
10 #include<queue>
11 #include<set>
12 #include<bitset>
13 #include<memory.h>
14 #include<time.h>
15 #include<stdio.h>
16 #include<stdlib.h>
17 #include<string.h>
18 //#include<stdbool.h>
19 #include<math.h>
20 #define min(a,b) ((a)<(b)?(a):(b))
21 #define max(a,b) ((a)>(b)?(a):(b))
22 #define abs(a) ((a)>0?(a):(-(a)))
23 #define lowbit(a) (a&(-a))
24 #define sqr(a) ((a)*(a))
25 #define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))
26 #define mem(a,b) memset(a,b,sizeof(a))
27 #define eps (1e-8)
28 #define J 10000
29 #define mod 1000000007
30 #define MAX 0x7f7f7f7f
31 #define PI 3.14159265358979323
32 #define N 44
33 using namespace std;
34 typedef long long LL;
35 int cas,cass;
36 int n,m,lll,ans;
37 LL aans;
38 int a[N][N];
39 int sum;
40 int f[N][N][2004];
41 void print()
42 {
43     int i,j,k;
44     for(i=0;i<=sum;i++)
45         printf("%d\n",f[n][m][i]);
46 }
47 int main()
48 {
49     #ifndef ONLINE_JUDGE
50 //    freopen("1.txt","r",stdin);
51 //    freopen("2.txt","w",stdout);
52     #endif
53     int i,j,k;
54 
55 //    for(scanf("%d",&cass);cass;cass--)
56     for(scanf("%d",&cas),cass=1;cass<=cas;cass++)
57 //    while(~scanf("%s",s+1))
58 //    while(~scanf("%d",&n))
59     {
60         mem(f,MAX);
61         sum=0;
62         scanf("%d%d",&n,&m);
63         for(i=1;i<=n;i++)
64             for(j=1;j<=m;j++)
65                 scanf("%d",&a[i][j]);
66         f[0][1][0]=0;
67         for(i=1;i<=n;i++)
68         {
69             for(j=1;j<=m;j++)
70             {
71                 for(k=a[i][j];k<=59*30;k++)
72                 {
73                     f[i][j][k]=min(f[i][j][k],f[i-1][j][k-a[i][j]]+sqr(a[i][j]));
74                     f[i][j][k]=min(f[i][j][k],f[i][j-1][k-a[i][j]]+sqr(a[i][j]));
75                 }
76             }
77         }
78         ans=MAX;
79         for(i=0;i<=59*30;i++)
80         { 
81             if(f[n][m][i]==f[0][0][0])continue;
82             ans=min(ans,(n+m-1)*f[n][m][i]-sqr(i));
83         }
84         printf("Case #%d: %d\n",cass,ans);
85     }
86     return 0;
87 }
88 /*
89 //
90 
91 //
92 */