Poj 2184 (dp)

3757 단어 동적 기획
Cow Exhibition
Time Limit: 1000MS
 
Memory Limit: 65536K
Total Submissions: 7795
 
Accepted: 2852
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
USACO 2003 Fall
간단한 dp,0-1 가방의 모형, 모든 아이템을 선택하거나 선택하지 않습니다.두 가지 가치의 합이 모두 마이너스라는 것을 보증하기 위해 일차 상태가 어떤 가치를 기록하는 합이라고 생각하면 dp[i][j]는 전 i종의 물품을 대표하고 첫 번째 가치가 j일 때 두 번째 가치의 최대치를 대표한다.세부 사항: 1.및 은(는) 음수가 될 수 있으므로 값은 INF를 추가합니다.2 상태 이동의 방정식에서 조건 판단을 잘 하고 수조가 경계를 넘지 않도록 주의해야 한다.3 스크롤 배열. 
#include
#include
#include
#include
#include
#include
#include
using namespace std;
const int maxn = 100 + 5;
const int maxm = 200000 + 5;
const int INF = 100000 + 1;

int dp[2][maxm];
int a[maxn],b[maxn];

int main(){
    int n;
    while(scanf("%d",&n) != EOF){
        for(int i = 1;i <= n;i++){
            scanf("%d%d",&a[i],&b[i]);
        }
        for(int i = 0;i < 2;i++){
            for(int j = 0;j < maxm;j++) dp[i][j] = -INF;
        }
        if(a[1] == 0){
            dp[1][INF] = max(0,b[1]);
        }
        else{
            dp[1][INF] = 0;
            dp[1][a[1]+INF] = b[1];
        }
        for(int i = 2;i <= n;i++){
            for(int j = -INF;j <= INF;j++){
                if(j-a[i]+INF >= maxm || dp[(i-1)&1][j-a[i]+INF] == -INF || j-a[i]+INF < 0){
                    dp[i&1][j+INF] = dp[(i-1)&1][j+INF];
                }
                else{
                    dp[i&1][j+INF] = max(dp[(i-1)&1][j-a[i]+INF]+b[i],dp[(i-1)&1][j+INF]);
                }
            }
        }
        int ans = 0;
        for(int i = 0;i <= INF;i++){
            if(dp[n&1][i+INF] >= 0) {
                ans = max(ans,dp[n&1][i+INF]+i);
            }
        }
        printf("%d
",ans); } return 0; }

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