Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Telephone poles are part of an outdated technology. Cell phones nowadays allow us to call whoever we want, wherever we want, independent of any wire. There is one problem - without a service tower nearby a cell phone is useless.

In the absence of hills and mountains, a service tower will provide coverage for a circular area. Instead of planning where to place the wires, a wireless telephone company has to plan where to build its service towers. Building the towers too far apart causes holes in the coverage and increases complaints. Building the towers too close to each other results in large areas covered by more than one service tower, which is redundant and inefficient.

International Cell Phone Company is developing a network strategy to determine the optimal placement of service towers. Since most customers have replaced their old wired home phones by cell phones, the strategy for planning service towers is therefore to cover as many homes of customers as possible.

The figure below shows the service areas for the five towers ICPC's strategic department planned to build this year. Tower 5 will serve 24 customers, 6 of which are also served by tower 4. Towers 1, 2 and 3 have a common service area containing 3 customers.

Shortly after the plans for these five towers had been published, higher management issued a stop on all tower building. Protesting customers forced them to weaken this decree, and allow the building of three towers out of the five planned. These three towers should serve as many customers as possible, of course. Finding the best possible choice for the towers to build is not trivial (the best choice in this case is towers 2, 4 and 5, serving 68 customers).

You must write a program to help ICPC choose which towers to build in cases like these. If several choices of towers serve the same number of customers, choices including tower 1 are preferable. If this rule still leaves room for more than one solution, solutions including tower 2 are preferable, and so on.

Input 

The input file contains several test cases. The first line of each test case contains two positive integers: the number n (n20) of towers planned, and the number of towers to be actually built. The next line contains n numbers, each giving the number of customers served by a planned tower. (The first number refers to the first tower, and so on.) No tower serves more than a million people. The next line contains the number m (m10) of common service areas. Each of the next m lines contains the description of a common service area. Such a line starts with the number t (t > 1) of towers for which this is a common service area, followed by the t numbers of the towers. The last number on the line is the number of customers in the common service area. The last line of the input file consists of the two integers `0 0'.

Output 

For each test case, display the number of customers served and the best choice for the location of the towers. Follow the format of the sample output. All numbers are left-justified and there are no blank spaces at the end of any line. Print a blank line after each test case.

Sample Input 

5  3                                                       
15 20 25 30 24                                             
5                                                           
2  1 2     7                                               
3  1 2 3   3                                               
2  2 3     2                                               
2  3 4     5                                                
2  4 5     6                                               
5  3                                                       
25 25 25 25 25                                             
4                                                           
2  1 2     5 
2  2 3     5 
2  3 4     5 
2  4 5     5 
5  3 
25 25 25 25 25 
0 
0 0

Sample Output 

Case Number  1 
Number of Customers: 68 
Locations recommended: 2 4 5 
  
Case Number  2 
Number of Customers: 75 
Locations recommended: 1 3 5 
  
Case Number  3 
Number of Customers: 75 
Locations recommended: 1 2 3

---

題目描述：

給定一些基地台的設置關係，以及交集的服務人數，然而只能建造特定的塔數，
問一組能服務最多的設置方法，相同時取字典順序最小的取法。

題目解法：

模擬。

#include <stdio.h>
#include <iostream>
#include <algorithm>
using namespace std;
int rmask[20] = {}, mw[20], c[20];
int mx, ret;
int n, m, r;
int i;
void dfs(int idx, int n, int m, int mask, int i) {
    if(idx == m) {
        int sum = 0;
        for(i = 0; i < n; i++)
            if((mask>>i)&1)
                sum += c[i];
        for(i = 0; i < r; i++) {
            int val = __builtin_popcount(mask&rmask[i]);
            if(val >= 2)
                sum -= mw[i]*(val-1);
        }
        if(sum > mx) {
            mx = sum;
            ret = mask;
        }
        return;
    }
    for(; i < n; i++)
        dfs(idx+1, n, m, mask|(1<<i), i+1);
}
int main() {
    int cases = 0;
    while(scanf("%d %d", &n, &m) == 2 && n) {
        for(i = 0; i < n; i++)
            scanf("%d", &c[i]);
        scanf("%d", &r);
        for(i = 0; i < r; i++) {
            rmask[i] = 0;
            int x, y;
            scanf("%d", &x);
            while(x--) {
                scanf("%d", &y);
                y--;
                rmask[i] |= 1<<y;
            }
            scanf("%d", &mw[i]);
        }
        mx = 0, ret = 0;
        dfs(0, n, m, 0, 0);
        printf("Case Number  %d\n", ++cases);
        printf("Number of Customers: %d\n", mx);
        printf("Locations recommended:");
        for(i = 0; i < n; i++)
            if((ret>>i)&1)
                printf(" %d", i+1);
        puts("\n");
    }
    return 0;
}