POJ--2184--Cow Exhibition--01背包
Cow Exhibition
Time Limit: 1000MS
Memory Limit: 65536K
Total Submissions: 9429
Accepted: 3624
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
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.
原諒我的愛國，所以靜不下心來看題，就直接在網上百度結題報告找題意，網上的基本都是模模糊糊，跟放屁一樣。
題意：S和F分別是牛的兩個屬性，告訴你N頭牛的這兩個屬性，然後要求你選出隨意數量的牛，使得所有牛的S和F值加起來最大，同時所有牛的S值加起來不小於零，F值加起來也不小於零
解析：當作01背包來做的話就是把其中一個屬性當作容量，這樣下來，一定容量下得到的價值取最大就好了。
比如S當作容量，dp[X]的含義就成了：S值為X的時候F值最大量
但是S值有小於零的，這時我們考慮一下最小有多小就好做了，自己想吧！
#include
#include
#include
#include
using namespace std;
int dp[222222]; //最小是-10W，那麼把整個數組留10W以上的空間給負數就好了
int main (void)
{
int n,m,i,j,k,l,a,b,x=111111; //x標記0的位置
int Max,Min; //最大最小邊界，方便遍歷已經求過值的范圍裡的數據
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<222222;i++)
{
dp[i]=-999999;
}
Max=Min=x;
dp[x]=0;
for(i=0;i=Min;j--) //正的要從大到小，01背包
{
if(dp[j+a]=0&&l
