POJ 2184 01背包+负数处理

Cow Exhibition

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 10200		Accepted: 3977

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 (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

Sample Output

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.

Source

USACO 2003 Fall

题目意思：

有n头牛，每头牛有自己的聪明值和幽默值，选出几头牛使得选出牛的聪明值总和大于0、幽默值总和大于0，求聪明值和幽默值总和相加最大为多少。

思路：

每头牛要么选要么不选，背包问题。以聪明值为数组维度，幽默值为数组值即构成01背包。但是聪明值可能有负数，所以需要增加数组长度。

也就是把坐标0向正方向移动：

0。。。mid。。。maxh

0-mid之间聪明值为负数，mid-maxh之间聪明值为正数，mid为0。

然后就可以进行01背包了。

注意，当聪明值小于0的时候背包是从小到大的，因为dp[i]=max(dp[i],dp[i-v]+w)；其中v为负数，那么就相当于从大的转移到小的，为防止重复则反过来for。

代码：

 #include <cstdio>

 #include <cstring>

 #include <algorithm>

 #include <iostream>

 #include <vector>

 #include <queue>

 #include <cmath>

 #include <set>

 using namespace std;

 #define N 200005

 #define sh 100000

 #define inf 999999999

 int max(int x,int y){return x>y?x:y;}

 int min(int x,int y){return x<y?x:y;}

 int abs(int x,int y){return x<?-x:x;}

 int dp[N];

 struct node{

     int v, w;

 }a[];

 int n;

 main()

 {

     int i, j, k;

     while(scanf("%d",&n)==){

         for(i=;i<n;i++) scanf("%d %d",&a[i].v,&a[i].w);

         for(i=;i<N;i++) dp[i]=-inf;

         dp[sh]=;

         for(i=;i<n;i++){

             if(a[i].v>){

                 for(k=N-;k>=a[i].v;k--) dp[k]=max(dp[k-a[i].v]+a[i].w,dp[k]);

             }

             else{

                 for(k=;k<N+a[i].v;k++) dp[k]=max(dp[k-a[i].v]+a[i].w,dp[k]);

             }

         }

         int ans=;

         for(i=sh;i<N;i++){

             if(dp[i]>) {

                 ans=max(ans,i-sh+dp[i]);

             }

         }

         printf("%d\n",ans);

     }

 }

巴特西

POJ 2184 01背包+负数处理

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