D - Patisserie ABC     列舉狀態 加 貪心

Time limit : 2sec / Memory limit : 1000MB

Score: 400 points

Problem Statement

Takahashi became a pastry chef and opened a shop La Confiserie d'ABC to celebrate AtCoder Beginner Contest 100.

The shop sells N kinds of cakes. Each kind of cake has three parameters "beauty", "tastiness" and "popularity". The i

Ringo has decided to have M pieces of cakes here. He will choose the set of cakes as follows:

• Do not have two or more pieces of the same kind of cake.
• Under the condition above, choose the set of cakes to maximize (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity).

Find the maximum possible value of (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) for the set of cakes that Ringo chooses.

Constraints

• N is an integer between 1 and 1 000 (inclusive).
• M is an integer between 0 and N
   (inclusive).
• xi,yi,zi (1≤i≤N) are integers between −10 000 000 000 and 10 000 000 000 (inclusive).

Input

Input is given from Standard Input in the following format:

N M
x1 y1 z1
x2 y2 z2
 :  :
xN yN zN

Output

Print the maximum possible value of (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) for the set of cakes that Ringo chooses.

Sample Input 1

Copy

5 3
3 1 4
1 5 9
2 6 5
3 5 8
9 7 9

Sample Output 1

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56

Consider having the 2-nd, 4-th and 5-th kinds of cakes. The total beauty, tastiness and popularity will be as follows:

• Beauty: 1+3+9=13
• Tastiness: 5+5+7=17
• Popularity: 9+8+9=26

The value (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) here is 13+17+26=56. This is the maximum value.

Sample Input 2

Copy

5 3
1 -2 3
-4 5 -6
7 -8 -9
-10 11 -12
13 -14 15

Sample Output 2

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54

Consider having the 1-st, 3-rd and 5-th kinds of cakes. The total beauty, tastiness and popularity will be as follows:

• Beauty: 1+7+13=21
• Tastiness: (−2)+(−8)+(−14)=−24
• Popularity: 3+(−9)+15=9

The value (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) here is 21+24+9=54. This is the maximum value.

Sample Input 3

Copy

10 5
10 -80 21
23 8 38
-94 28 11
-26 -2 18
-69 72 79
-26 -86 -54
-72 -50 59
21 65 -32
40 -94 87
-62 18 82

Sample Output 3

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638

If we have the 3-rd, 4-th, 5-th, 7-th and 10-th kinds of cakes, the total beauty, tastiness and popularity will be −323, 66 and 249, respectively. The value (the absolute value of the total beauty) + (the absolute value of the total tastiness) + (the absolute value of the total popularity) here is 323+66+249=638. This is the maximum value.

Sample Input 4

Copy

3 2
2000000000 -9000000000 4000000000
7000000000 -5000000000 3000000000
6000000000 -1000000000 8000000000

Sample Output 4

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30000000000

The values of the beauty, tastiness and popularity of the cakes and the value to be printed may not fit into 32-bit integers.

給你n個蛋糕，我們可以從中選擇m個，每個蛋糕都有三列價值（正負都有），先讓你求選擇m個蛋糕的最大價值，

即豎列的價值相加，如果有負數，就取絕對值，得到的和再相加。

思路：只有1000個蛋糕，每個蛋糕的三個價值也只有八種狀態，就是

---    +++   --+    -++    -+-     +--     ++-     +-+

所以對於每個蛋糕我們可以列舉他的價值貢獻的可能，然後取一個最大值，接著對n個蛋糕的總價值

排序一遍，取前m個即可

程式碼如下：

#include <bits/stdc++.h>

using namespace std;

const int maxn = 2000;
struct node {
    long long a,b,c,s;
} g[maxn];
int n,m;
long long s[maxn];
long long calc(int flag1,int flag2,int flag3,int i ) {
    long long ans = 0;
    if(flag1) ans += g[i].a;
    else ans -= g[i].a;
    if(flag2) ans += g[i].b;
    else ans -= g[i].b;
    if(flag3) ans += g[i].c;
    else ans -= g[i].c;
    return ans;
}

bool cmp(long long a,long long b) {
    return a > b;
}

long long getMax(int a,int b,int c) {
    for(int i = 1; i <= n; i++)
        s[i] = calc(a,b,c,i);
    sort(s + 1,s + 1 + n,cmp);
    long long sum = 0;
    for(int i = 1; i <= m; i++)
        sum += s[i];
    return sum;
}
int main() {
    while(cin >> n >> m) {
        memset(s,0,sizeof(s));
        long long ans = 0;
        for(int i = 1; i <= n; i++) cin >> g[i].a >> g[i].b >> g[i].c,g[i].s = g[i].a + g[i].b + g[i].c;
        ans = max(ans,getMax(1,1,1));
        ans = max(ans,getMax(1,1,0));
        ans = max(ans,getMax(1,0,1));
        ans = max(ans,getMax(0,1,1));
        ans = max(ans,getMax(1,0,0));
        ans = max(ans,getMax(0,0,1));
        ans = max(ans,getMax(0,1,0));
        ans = max(ans,getMax(0,0,0));
        cout << ans << endl;
    }
    return 0;
}