台州 OJ 5072 Cow Exhibition 01背包

#include <iostream>
#include <cstring>
using namespace std;

const int MAX = 2005;
const int INF = 0x3f3f3f3f;
const int zero = 100000; 

int n;
int now[MAX*100], pre[MAX*100];
int s[MAX], f[MAX];

int main(){
    freopen("input.txt", "r", stdin);
    int ans = 0, minS = 0, maxS = 0;
    cin >> n;
    for(int i=1; i<=n; i++){
        cin >> s[i] >> f[i];
        if(s[i] >= 0 && f[i] >= 0){        //smartness 和 funness 都大于 0 的牛一定选 
            ans += s[i] + f[i];
            i--; n--;
        }else if(s[i] < 0 && f[i] < 0){        //都小于 0 的一定不选 
            i--; n--;
        }
    }
    
    //初始化 
    memset(now, -INF, sizeof(now));
    now[zero] = 0;
    memcpy(pre, now, sizeof(now));
    
    //DP
    minS = maxS = zero;        // 0 变成 100000，将 smartness 为负数的情况转化为全部是正数 
    for(int i=1; i<=n; i++){
        for(int j=minS; j<=maxS; j++){
            now[j+s[i]] = max(pre[j+s[i]], pre[j] + f[i]);    //smartness j+s[i] 时，funness 的最大值，考虑第 i 头牛放不放 
        }
        minS = min(minS + s[i], minS);
        maxS = max(maxS + s[i], maxS);
        memcpy(pre, now, sizeof(now));
    }
    
    int maxn = 0;
    for(int i=zero; i<=maxS; i++){
        if(now[i] >= 0){
            maxn = max(maxn, (i-zero) + now[i]);    //要 - zero 转化为原来的 smartness 
        }
    }
    
    cout << ans + maxn;
    
    return 0;
}