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

Citizen attention offices

The Mayor of a city wants to improve the attentions of the local government to the citizens. For that, five offices to attend to consults will be open in the city. The Mayor thinks that the attentions would be better if the offices are close to where the people live. To decide where to open the offices, the city has been divided in 25 areas, in a square:

0	1	2	3	4
5	6	7	8	9
10	11	12	13	14
15	16	17	18	19
20	21	22	23	24

and the quantity of people living in each square is counted, thus assigning to each area the number (an integer number) of thousands of people living there. For example:

0	0	2	3	1
5	2	2	2	0
1	1	2	3	0
0	0	2	1	5
2	1	2	0	4

The number of people in each area is less or equal to 10 million.

For each area, the minimum distance to the five offices is obtained as follows:  the distance from one area to an office is obtained  as the number of movements to go from the area to the area where the office will be installed (the movements are in horizontal and vertical; for example, from (1,1) to (2,3) three movements are necessary). And we want to minimize the sum of the minimum distances from all the areas, but having into account the quantity of people living in each area. Thus, for each area the distance to an office is multiplied by the value associated to the area.

The Input

The input begins with a line where the number of test cases (t) is indicated. The data for each test case appear in sucessive lines. In each test case the first line contains the number of areas in the city with non null population (n) and in each one of the n following lines three numbers appear, the first and the second indicate the row and the column of the area (they are numbered from 0 to 4) and the third the number of thousand of people living in the area.

The Output

The output consists of a line for each problem, with five numbers with a space between each two numbers, and no space after the last number. The numbers in the solutions are written in increasing order. If there is more than one optimum solution, you must output the solution with lowest first value. If the first value coincides, you must output the solution with lowest second value, and so on.

Sample Input

4
1
2 2 1
4
0 0 1
4 4 1
0 4 1
4 0 1
5
0 0 1
1 1 1
2 2 1
3 3 1
4 4 1
7
4 2 2
3 3 1
2 4 3
2 1 1
1 3 4
1 2 2
1 0 1

Sample Output

0 1 2 3 12
0 1 4 20 24
0 6 12 18 24
5 7 8 14 22

---

題目意思 : 將 25 個房間的人移動到 5 個房間, 求最小的步數
並輸出序列最小的最佳解

做法 : 窮舉所有組合
想不到什麼好的剪枝方法, 原本想用 A* 去做, 可是這樣要用bitset 去做狀態記錄, 這樣就有點煩人了。 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define oo 0xfffffff
int grid[5][5], ch[5], ans[5];
int minv;
void dfs(int idx, int i) {
    if(idx == 5) {
        int sum = 0, tmp;
        int i, j, k, x, y;
        for(i = 0; i < 5; i++) {
            for(j = 0; j < 5; j++) {
                if(grid[i][j]) {
                    tmp = oo;
                    for(k = 0; k < 5; k++) {
                        x = ch[k]/5, y = ch[k]%5;
                        if(abs(x-i)+abs(y-j) < tmp)
                            tmp = abs(x-i)+abs(y-j);
                    }
                    sum += grid[i][j]*tmp;
                }
            }
        }
        if(sum < minv) {
            minv = sum;
            for(i = 0; i < 5; i++)
                ans[i] = ch[i];
        }
        return;
    }
    for(; i < 25; i++) {
        ch[idx] = i;
        dfs(idx+1, i+1);
    }
}
int main() {
    int t, n, i, j, k;
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        memset(grid, 0, sizeof(grid));
        while(n--) {
            scanf("%d %d %d", &i, &j, &k);
            grid[i][j] += k;
        }
        minv = oo;
        dfs(0, 0);
        for(i = 0; i < 5; i++) {
            if(i)
                putchar(' ');
            printf("%d", ans[i]);
        }
        puts("");
    }
    return 0;
}