Poj 2184 (dp)

Cow Exhibition

Time Limit: 1000MS Memory Limit: 65536KTotal 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 78 -66 -32 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

简单dp，0-1背包的模型，每个物品，选或者不选。要保证两种价值的和都是负数，所以想到有一维状态是记录某种价值的和，那么dp[i][j]就代表的是前i种物品，第一种价值为j时，第二种价值的最大值。细节处理：1. 和可以为负数，所以值都加上一个INF。2 状态转移的方程中，注意写好条件判断，数组不要越界了。3 滚动数组。 

#include<cstdio>#include<iostream>#include<map>#include<vector>#include<algorithm>#include<set>#include<cstring>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\n",ans);    }    return 0;}