POJ2184 Cow Exhibition 【01背包】

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 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

题意：有n头牛，每头牛有两个属性s,f,从中选择一些牛使得这些牛的属性s的和TS为非负，属性f的和TF也为非负。并且求TS+TF的最大值。

题解：dp[i]中i表示TS为i时能得到的TF最大值。具体的解法也是参考网上的解题报告，Click

#include <stdio.h>
#include <string.h>
#include <algorithm>

#define maxn 200010
#define delta 1000 * 100
#define inf 0x3f3f3f3f
using namespace std;

int dp[maxn], N, ans; // dp[i] stands when TS is i, the max TF
struct Node {
    int s, f;
} arr[102];

int main() {
    // freopen("stdin.txt", "r", stdin);
    int i, j, s, f, begin, end, u, v, step;
    scanf("%d", &N);
    for(i = 0; i < N; ++i) {
        scanf("%d%d", &s, &f);
        if(s <= 0 && f <= 0) {
            --i; --N;
            continue;
        }
        arr[i].s = s;
        arr[i].f = f;
    }
    fill(dp, dp + maxn, -inf);

    u = v = 0; dp[0 + delta] = 0;
    for(i = 0; i < N; ++i) {
        step = 1;
        u = min(u, u + arr[i].s);
        v = max(v, v + arr[i].s);

        begin = u; end = v;
        if(arr[i].s > 0) {
            step = -1;
            swap(begin, end);
        }

        for(j = begin; j != end + step; j += step) // j maybe negative
            dp[j + delta] = max(dp[j+delta], dp[j+delta-arr[i].s] + arr[i].f);
    }

    for(i = 0; i <= v; ++i)
        if(dp[i+delta] >= 0)
            ans = max(ans, dp[i+delta] + i);
    printf("%d\n", ans);
    return 0;
}