Cow Exhibition POJ 2184(01背包dp)

Posted 2022-08-30 vbel

原题

题目链接

题目分析

明显的01背包,但还是由细节需要处理,设置dp[i][j]=前i头牛中选的TF为j时最大的TS值,由于TF可能为负数因此要加一个基数使其在大于等于零,dp初始化为-1,dp[0][0]=0.更新的时候dp[i][j]=max(dp[i-1][j-f[i]]+s[i],dp[i-1][j]).观察这个dp的更新可以发现能状态压缩,当f[i]>=0时,j从大到小更新,当f[i]<0时,j从小到大更新.最后从基数往上扫dp[j]+j-基数的最大值就是答案,注意前提是dp[i]>=0.

代码

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <utility>
 4 #include <cstdio>
 5 #include <cmath>
 6 #include <cstring>
 7 #include <string>
 8 #include <vector>
 9 #include <stack>
10 #include <queue>
11 #include <map>
12 #include <set>
13 
14 using namespace std;
15 typedef long long LL;
16 const int INF_INT=0x3f3f3f3f;
17 const LL INF_LL=0x3f3f3f3f3f3f3f3f;
18 
19 struct N
20 
21     int s,f;
22 cow[200];
23 int dp[300000];
24 
25 bool cmp(N a,N b)
26 
27     return a.s>b.s;
28 
29 
30 int main()
31 
32 //    freopen("black.in","r",stdin);
33 //    freopen("black.out","w",stdout);
34     int n;
35     cin>>n;
36     int uplimit=0,downlimit=0;
37     for(int i=1;i<=n;i++)
38     
39         cin>>cow[i].s>>cow[i].f;
40         if(cow[i].f>0) uplimit+=cow[i].f;
41         else downlimit+=cow[i].f;
42     
43     sort(cow+1,cow+n+1,cmp);
44     int limit=max(uplimit,abs(downlimit));
45 /*    printf("limit=%d\n",limit);
46     for(int i=1;i<=n;i++) printf("cow %d s %d f %d\n",i,cow[i].s,cow[i].f);*/
47     for(int j=0;j<=2*limit;j++) dp[j]=-1*INF_INT;
48     dp[limit]=0;
49     for(int i=1;i<=n;i++)
50     
51         if(cow[i].f>=0)
52         
53             for(int j=2*limit;j>=0;j--)
54             
55                 int from=j-cow[i].f;
56                 if(from>=0&&from<=limit*2&&dp[from]!=-1*INF_INT)
57                 
58                     dp[j]=max(dp[j],dp[from]+cow[i].s);
59                 
60             
61         
62         else
63         
64             for(int j=0;j<=2*limit;j++)
65             
66                 int from=j-cow[i].f;
67                 if(from>=0&&from<=limit*2&&dp[from]!=-1*INF_INT)
68                 
69                     dp[j]=max(dp[j],dp[from]+cow[i].s);
70                 
71             
72         
73   //      for(int j=0;j<=2*limit;j++) if(dp[now][j]!=-1) printf("dp %d %d=%d\n",i,j-limit,dp[now][j]);
74     
75     int ans=0;
76     for(int i=limit;i<=limit*2;i++) if(dp[i]>=0) ans=max(ans,dp[i]+i-limit);
77     cout<<ans<<endl;
78     return 0;
79