poj 2184 Cow Exhibition（dp之01背包变形）

Description

"Fat and docile, big and dumb, they look so stupid, they aren't much
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 ( <= N <= ) cows a thorough interview and determined two values for each cow: the smartness Si (- <= Si <= ) of the cow and the funness Fi (- <= Fi <= ) 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 : A single integer N, the number of cows 

* Lines ..N+: Two space-separated integers Si and Fi, respectively the smartness and funness for each cow. 

Output

* Line : 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 . 

Sample Input

-
 -
 -

- -

Sample Output

Hint

OUTPUT DETAILS: 

Bessie chooses cows , , and , giving values of TS = -++ =  and TF
= -+ = , so + = . Note that adding cow  would improve the value
of TS+TF to , but the new value of TF would be negative, so it is not
allowed. 

Source

USACO 2003 Fall

 

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<cmath>
 #include<stdlib.h>
 using namespace std;
 #define inf 1<<30
 #define N 106
 int dp[];
 int a[N],b[N];
 int main()
 {
     int n;
     while(scanf("%d",&n)==){
         for(int i=;i<n;i++){
             scanf("%d%d",&a[i],&b[i]);
         }
         //memset(dp,inf,sizeof(dp));
         for(int i=;i<;i++){
             dp[i]=-inf;
         }
         dp[]=;
         for(int i=;i<n;i++){
             //if(a[i]<0 && b[i]<0) continue;
             if(a[i]>){
                 for(int j=;j>=a[i];j--){
                     dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
                 }
             }
             else{
                 for(int j=a[i];j<=+a[i];j++){
                     dp[j]=max(dp[j],dp[j-a[i]]+b[i]);
                 }
             }
         }
         int ans=;
         for(int i=;i<=;i++){
             if(dp[i]>=){
                 ans=max(ans,dp[i]+i-);
             }
         }
         printf("%d\n",ans);
     }
     return ;
 }

 

题意：每行给出si和fi，代表牛的两个属性，然后要求选出几头牛，是的则求出总S与总F的和，注意S与F都不能为负数

思路：很明显的就是取与不取的问题，对于这类问题的第一想法就是背包，但是这道题目很明显与一般的背包不同，因为有负数，但是联想到以前也有这种将负数存入下标的情况，那就是将数组开大，换一种存法

我们用dp[i]存放每个s[i]能得到的最佳F，那么我们就可以根据s[i]的取值采取两种不同的01背包取法，在取完之后，然后再根据背包的有无再去求得最佳答案即可

 

附上大神代码：

 #include <stdio.h>
 #include <string.h>
 #include <algorithm>
 using namespace std;

 int dp[];
 const int inf = <<;

 int main()
 {
     int n,s[],f[],i,j,ans;
     while(~scanf("%d",&n))
     {
         for(i = ; i<=; i++)
             dp[i] = -inf;
         dp[] = ;
         for(i = ; i<=n; i++)
             scanf("%d%d",&s[i],&f[i]);
         for(i = ; i<=n; i++)
         {
             if(s[i]< && f[i]<)
                 continue;
             if(s[i]>)
             {
                 for(j = ; j>=s[i]; j--)//如果s[i]为整数，那么我们就从大的往小的方向进行背包
                     if(dp[j-s[i]]>-inf)
                         dp[j] = max(dp[j],dp[j-s[i]]+f[i]);
             }
             else
             {
                 for(j = s[i]; j<=+s[i]; j++)//为负数则需要反过来
                     if(dp[j-s[i]]>-inf)
                         dp[j] = max(dp[j],dp[j-s[i]]+f[i]);
             }
         }
         ans = -inf;
         for(i = ; i<=; i++)//因为区间100000~200000才是表示的整数，那么此时的i就是之前背包中的s[i]，如果此时dp[i]也就是f[i]大于等于0的话，我们再加上s[i]（此时为i），然后减去作为界限的100000，就可以得到答案
         {
             if(dp[i]>=)
                 ans = max(ans,dp[i]+i-);
         }
         printf("%d\n",ans);
     }

     return ;
 }