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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.
Source
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<cmath> 6 #include<stdlib.h> 7 using namespace std; 8 #define inf 1<<30 9 #define N 106 10 int dp[200006]; 11 int a[N],b[N]; 12 int main() 13 { 14 int n; 15 while(scanf("%d",&n)==1){ 16 for(int i=0;i<n;i++){ 17 scanf("%d%d",&a[i],&b[i]); 18 } 19 //memset(dp,inf,sizeof(dp)); 20 for(int i=0;i<200006;i++){ 21 dp[i]=-inf; 22 } 23 dp[100000]=0; 24 for(int i=0;i<n;i++){ 25 //if(a[i]<0 && b[i]<0) continue; 26 if(a[i]>0){ 27 for(int j=200000;j>=a[i];j--){ 28 dp[j]=max(dp[j],dp[j-a[i]]+b[i]); 29 } 30 } 31 else{ 32 for(int j=a[i];j<=200000+a[i];j++){ 33 dp[j]=max(dp[j],dp[j-a[i]]+b[i]); 34 } 35 } 36 } 37 int ans=0; 38 for(int i=100000;i<=200000;i++){ 39 if(dp[i]>=0){ 40 ans=max(ans,dp[i]+i-100000); 41 } 42 } 43 printf("%d\n",ans); 44 } 45 return 0; 46 }
1 #include <stdio.h> 2 #include <string.h> 3 #include <algorithm> 4 using namespace std; 5 6 int dp[200005]; 7 const int inf = 1<<30; 8 9 int main() 10 { 11 int n,s[200],f[200],i,j,ans; 12 while(~scanf("%d",&n)) 13 { 14 for(i = 0; i<=200000; i++) 15 dp[i] = -inf; 16 dp[100000] = 0; 17 for(i = 1; i<=n; i++) 18 scanf("%d%d",&s[i],&f[i]); 19 for(i = 1; i<=n; i++) 20 { 21 if(s[i]<0 && f[i]<0) 22 continue; 23 if(s[i]>0) 24 { 25 for(j = 200000; 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 26 if(dp[j-s[i]]>-inf) 27 dp[j] = max(dp[j],dp[j-s[i]]+f[i]); 28 } 29 else 30 { 31 for(j = s[i]; j<=200000+s[i]; j++)//为负数则需要反过来 32 if(dp[j-s[i]]>-inf) 33 dp[j] = max(dp[j],dp[j-s[i]]+f[i]); 34 } 35 } 36 ans = -inf; 37 for(i = 100000; i<=200000; 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 38 { 39 if(dp[i]>=0) 40 ans = max(ans,dp[i]+i-100000); 41 } 42 printf("%d\n",ans); 43 } 44 45 return 0; 46 }
poj 2184 Cow Exhibition（dp之01背包变形）
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原文地址：http://www.cnblogs.com/UniqueColor/p/4780849.html
