poj 2184 01背包變形【背包dp】
阿新 • • 發佈：2018-02-04
accept amp 參考 www. limit believe 另一個 gun like
fun..."
- Cows with Guns by Dana Lyons
The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow.
Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si‘s and, likewise, the total funness TF of the group is the sum of the Fi‘s. Bessie wants to maximize the sum of TS and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative.
* Lines 2..N+1: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow.
Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF
= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the value
of TS+TF to 10, but the new value of TF would be negative, so it is not
allowed.
POJ 2184
Cow Exhibition| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 14657 | Accepted: 5950 |
Description
"Fat and docile, big and dumb, they look so stupid, they aren‘t muchfun..."
- Cows with Guns by Dana Lyons
The cows want to prove to the public that they are both smart and fun. In order to do this, Bessie has organized an exhibition that will be put on by the cows. She has given each of the N (1 <= N <= 100) cows a thorough interview and determined two values for each cow: the smartness Si (-1000 <= Si <= 1000) of the cow and the funness Fi (-1000 <= Fi <= 1000) of the cow.
Bessie must choose which cows she wants to bring to her exhibition. She believes that the total smartness TS of the group is the sum of the Si‘s and, likewise, the total funness TF of the group is the sum of the Fi‘s. Bessie wants to maximize the sum of TS and TF, but she also wants both of these values to be non-negative (since she must also show that the cows are well-rounded; a negative TS or TF would ruin this). Help Bessie maximize the sum of TS and TF without letting either of these values become negative.
Input
* Line 1: A single integer N, the number of cows* Lines 2..N+1: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow.
Output
* Line 1: One integer: the optimal sum of TS and TF such that both TS and TF are non-negative. If no subset of the cows has non-negative TS and non- negative TF, print 0.Sample Input
5
-5 7
8 -6
6 -3
2 1
-8 -5
Sample Output
8
Hint
OUTPUT DETAILS:Bessie chooses cows 1, 3, and 4, giving values of TS = -5+6+2 = 3 and TF
= 7-3+1 = 5, so 3+5 = 8. Note that adding cow 2 would improve the value
of TS+TF to 10, but the new value of TF would be negative, so it is not
allowed.
題意：給定N頭牛的smart值和fun值，求選出若幹頭牛使得smart值加fun值之和最大，且兩值分別之和不能小於零。
題解：每一頭牛都是選或不選，做法應該是01背包dp。但是空間容量沒有，有兩個值s[i],f[i]。把其中一個值作為背包容量，另一個作為價值。定義i表示前k頭牛的智力總和，dp[i]存儲前k頭牛智力總和為i時的幽默值總和的最大值。由於s[i]可能為負，且s[i]代表“空間”，所以s[i]為負時要像完全背包一樣正序循環。理由見參考博客。且根據數據範圍易知∑s[i]屬於[-1e5,1e5]，因為“空間”有的部分，所以可以整體平移1e5，使得“空間”大小非負，那麽此時原來的0點就變成了1e5，空間容量最大就為2e5。這樣求得dp[i]表示幽默值總和，i表示智力總和，又因為有平移。所以最後結果為ans=max(dp[i]+i-1e5),dp[i]>0.
下面代碼C++編譯器能過，G++ runtime error ,我還特意把數組開大了，不是很懂。。。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int maxn=1e5; int dp[2*maxn+5000]; int s[105],f[105]; int main() { int N; cin>>N; for(int i=0;i<=2*maxn+1020;i++) dp[i]=-1e8; dp[maxn]=0; for(int i=1;i<=N;i++) cin>>s[i]>>f[i]; for(int i=1;i<=N;i++){ if(s[i]>=0){ for(int j=maxn*2;j>=s[i];j--) dp[j]=max(dp[j],dp[j-s[i]]+f[i]); } else { for(int j=s[i];j<=maxn*2;j++) dp[j]=max(dp[j],dp[j-s[i]]+f[i]); } } int ans=0; for(int i=maxn;i<=2*maxn;i++) if(dp[i]>0) ans=max(ans,dp[i]+i-maxn); cout<<ans<<endl; return 0; }
這份仿照別人的寫的：
#include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<vector> using namespace std; #define S 100000 #define M 200000 #define INF 0x3f3f3f3f int dp[M+5]; int main() { int n,m; int a[105],b[105]; while(cin>>n) { for(int i=0;i<n;i++) cin>>a[i]>>b[i]; memset(dp,-INF,sizeof(dp)); dp[S]=0; for(int i=0;i<n;i++) { if(a[i]>0){ for(int j=M;j>=a[i];j--) //if(dp[j-a[i]]>-INF) dp[j]=max(dp[j],dp[j-a[i]]+b[i]); } else { for(int j=0;j-a[i]<=M;j++) //if(dp[j-a[i]]>-INF) dp[j]=max(dp[j],dp[j-a[i]]+b[i]); } } int ans=0; for(int i=S;i<=M;i++) if(dp[i]>=0) ans=max(ans,dp[i]+i-S); cout<<ans<<endl; } return 0; }
參考博客（感謝博主們）：
【1】：https://www.cnblogs.com/Findxiaoxun/p/3398075.html
【2】：https://www.cnblogs.com/pshw/p/5031109.html
【3】：http://blog.csdn.net/keysona/article/details/45751903
poj 2184 01背包變形【背包dp】