poj 2184 Cow Exhibition

/*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 Input5-5 78 -66 -32 1-8 -5Sample Output8思路：往01背包上靠，可是这每个物品竟然有两个价值，难道是二维费用背包，可是又没有确定下来的背包的容量。纠结纠结又纠结之后，队友传来消息：能不能用其中一个价值当做物品花费，另一个价值当做真正的价值，这样做01背包，把背包容量不定，更新所有的0-100*2000，最后输出最大的那个结果。整理起来就是用TS做物品花费，TF做物品价值，做01背包。但是要注意有负数的存在，如果用一维数组做背包，一定要注意第二重循环的顺序（因为背包的结果是用前一个状态转移来的，更新当前一层的数据时不能更改上一层的数据）。如果是二维数组，就不用纠结循环的顺序，不过要更新小于当前物品花费的数据。*/#include <stdio.h>#include <string.h>#define max(a,b) a>b?a:bint main(){int i,j,s[2005],f[2005],dp[200005];int n;while(scanf("%d",&n)!=EOF){for(i=0;i<n;i++){scanf("%d%d",&s[i],&f[i]);}memset(dp,-0x3f3f3f3f,sizeof(dp));dp[100000]=0;for(i=0;i<n;i++){if(s[i]<0&&f[i]<0)continue;else if(s[i]<=0){for(j=s[i];j<=200000+s[i];j++){dp[j]=max(dp[j],dp[j-s[i]]+f[i]);}}else{for(j=200000;j>=s[i];j--){dp[j]=max(dp[j],dp[j-s[i]]+f[i]);}}}int ans=-0x3f3f3f3f;for(i=100000;i<200000;i++){if(dp[i]>=0)ans=max(ans,dp[i]+i-100000);}printf("%d\n",ans); }  return 0;}