Equipment UVA - 1508(子集补集)

The Korea Defense and Science Institute, shortly KDSI, has been putting constant effort into new
equipment for individual soldiers for recent years, and at last released N new types of equipment.
KDSI has already done evaluation of each of the N types of equipment, finally resulting in scores in five
categories: attack improvement, defense improvement, vision improvement, portability, and easiness of
usage. The score in each category is quantified as an integer in the range 0 to 10,000, and the rating
of each type of equipment is thus represented as a sequence of five integers.
In consideration of costs and capability of average individual soldiers, KDSI also reported that each
soldier will be able to install at most K types of equipment on his body to extend his ability. If a single
type is installed on a soldier, then his ability in each category is extended by the specified score of that
type. Moreover, if a soldier installs more than one type of equipment, then his ability in each category
is extended by the maximum score of the chosen types in that category. For example, if the vision
improvement scores of type a and type b are 10 and 15, respectively, then installing a combination
of two types a and b will result in a vision improvement by their maximum score 15. We call the
maximum score 15 the extension score of a category; so, the extension score of vision improvement for
combination {a, b} is 15 in this example.
KDSI now started devising a way of finding an optimal combination of K types of equipment for best
performance of individual soldiers. While a force can sometimes be of a special purpose so that a certain
category would be more important than the others, every single category is, however, regarded equally
important in general. For this general purpose, KDSI defined the objective score of a combination of
equipment to be the sum of the extension scores of the five categories for the combination. KDSI thus
wants to find a best combination of K types of equipment such that its objective score is maximized
among all possible combinations of K types. You are asked by KDSI to devise and write a computer
program that finds the objective score of a best combination of K types of equipment, that is, the
maximum possible objective score for all possible combinations of K types among the given N types of
equipment.
Put differently, you are given N types of equipment {1, . . . , N} and their ratings Ri represented by
five integers Ri = (ri,1, ri,2, ri,3, ri,4, ri,5) with 0 ≤ ri,j ≤ 10, 000 for each i = 1, . . . , N and j = 1, . . . , 5.
Given another natural number K (1 ≤ K ≤ N), your program has to compute the objective score of a
best combination of K types of equipment.
For example, consider an input instance in which N = 4, K = 2, and each Ri
is given as below:
R1 = (30, 30, 30, 30, 0)
R2 = (50, 0, 0, 0, 0)
R3 = (0, 50, 0, 50, 10)
R4 = (0, 0, 50, 0, 20).
Then, choosing R1 and R3 forms a best combination of two types {1, 3} and yields the objective
score 30 + 50 + 30 + 50 + 10 = 170, which will be the answer of a correct program.
Input
Your program is to read from standard input. The input consists of T test cases. The number T of
test cases is given in the first line of the input. From the second line, each test case is given in order,
consisting of the following: a test case contains two integers N (1 ≤ N ≤ 10, 000) and K (1 ≤ K ≤ N)
in its first line, and is followed by N lines each of which consists of five integers inclusively between
0 and 10,000, representing the five scores ri,1, ri,2, ri,3, ri,4, and ri,5 of each type i of equipment for
i = 1, . . . , N in order. Two consecutive integers in one line are separated by a single space and there is
no empty line between two consecutive test cases.
Output
Your program is to write to standard output. Print exactly one line for each test case. The line should
contain a single integer that is the objective score of a best combination of K types of equipment; the
maximum possible objective score for all possible combinations of K types among the given N types
of equipment for the corresponding test case.
The following shows sample input and output for two test cases.
Sample Input
2
4 2
30 30 30 30 0
50 0 0 0 0
0 50 0 50 10
0 0 50 0 20
5 1
10 20 60 0 0
0 0 20 50 30
30 50 20 20 0
10 10 10 20 30
30 0 20 10 20
Sample Output
170
120

给出 n 组数据,每组数据是一个五元组,问当你从这 n 组数据中取出 m 组时,可以获得的最大价值是多少。

根据题意,当 m>=5 时,可以取上每一列的最大值,这样获得的价值是最大的。

当m < 5 时,如果对 n 去dfs,虽然m只有4,但是复杂度还是太高了....没莽过去...

那么考虑对五元组,对于一个五元组一共只有 31 种取法,打表出这 31 种取法可以取得的最大值。(mx[i]是 i 状态下只取一组数据的最大值)

接下来对于全集31,开始 dfs。(1表示还可以取,0表示已经取了)

对于每次开始 dfs 的状态 u,枚举他的子集 i,以及对补集 i^u 进行下一步的dfs。

这时候还有一种情况,两步 dfs 取的子集,事实上都是由同一组数据来的,那么这种应该要排除。

但是其实这种情况一定不可能是最优解,因为这两个不同子集都是由同一个数据来的,那么这两个子集的并集的最优解,一定也是从这组数据来的,所以只需要通过一步来完成,那么通过上面那种方法,必然会浪费一步,从而使最后的答案变小,所以一定会被排除。

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*/
#include <map>
#include <set>
#include <list>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <string>
#include <vector>
#include <cstdio>
#include <bitset>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define  lowbit(x)  x & (-x)
#define  mes(a, b)  memset(a, b, sizeof a)
#define  fi         first
#define  se         second
#define  pii        pair<int, int>
#define  INOPEN     freopen("in.txt", "r", stdin)
#define  OUTOPEN    freopen("out.txt", "w", stdout)

typedef unsigned long long int ull;
typedef long long int ll;
const int    maxn = 1e4 + 10;
const int    maxm = 1e5 + 10;
const ll     mod  = 1e9 + 7;
const ll     INF  = 1e18 + 100;
const int    inf  = 0x3f3f3f3f;
const double pi   = acos(-1.0);
const double eps  = 1e-8;
using namespace std;

int n, m;
int cas, tol, T;

struct Node{
    int a[6];
} node[maxn];
int mx[50];
int ans;

void dfs(int u, int cnt, int sum) {
    if(cnt == m) {
        ans = max(ans, sum);
        return ;
    }
    for(int i=u; i; i=(i-1)&u) {
        dfs(i^u, cnt+1, sum + mx[i]);
    }
}

int main() {
    scanf("%d", &T);
    while(T--) {
        mes(mx, 0);
        mes(node, 0);
        scanf("%d%d", &n, &m);
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=5; j++) {
                scanf("%d", &node[i].a[j]);
            }
        }
        if(m >= 5) {
            int r1, r2, r3, r4, r5;
            r1 = r2 = r3 = r4 = r5 = 0;
            for(int i=1; i<=n; i++) {
                for(int j=1; j<=5; j++) {
                    r1 = max(r1, node[i].a[1]);
                    r2 = max(r2, node[i].a[2]);
                    r3 = max(r3, node[i].a[3]);
                    r4 = max(r4, node[i].a[4]);
                    r5 = max(r5, node[i].a[5]);
                }
            }
            printf("%d\n", r1+r2+r3+r4+r5);
            continue;
        } else {
            for(int i=1; i<=n; i++) {
                for(int u=0; u<=31; u++) {
                    int sum = 0;
                    for(int j=1; j<=5; j++) {
                        if(u & (1<<(j-1))) {
                            sum += node[i].a[j];
                        }
                    }
                    mx[u] = max(mx[u], sum);
                }
            }
            ans = 0;
            dfs(31, 0, 0);
            printf("%d\n", ans);
        }
    }
    return 0;
}
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转载自www.cnblogs.com/H-Riven/p/10296887.html

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