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

Problem B: Nuts for nuts..

So as Ryan and Larry decided that they don't really taste so good, they realized that there are some nuts located in certain places of the island.. and they love them! Since they're lazy, but greedy, they want to know the shortest tour that they can use to gather every single nut!

Can you help them?

Input

You'll be given x, and y, both less than 20, followed by x lines of y characters each as a map of the area, consisting sorely of ".", "#", and "L". Larry and Ryan are currently located in "L", and the nuts are represented by "#". They can travel in all 8 adjacent direction in one step. See below for an example. There will be at most 15 places where there are nuts, and "L" will only appear once.

Output

On each line, output the minimum amount of steps starting from "L", gather all the nuts, and back to "L".

Sample Input

5 5
L....
#....
#....
.....
#....
5 5
L....
#....
#....
.....
#....

Sample Output

8
8

---

TSP 問題

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define oo 32767
#define min(x, y) ((x) < (y) ? (x) : (y))

int g[20][20], dp[1<<18][18], t2[1<<18];
void tsp(int state, int last) {
    if(dp[state][last] != oo)   return;
    int i, j, tmp;
    dp[state][last]--;
    for(i = state; i; i -= j) {
        j = i&(-i);
        tsp(state-j, t2[j]);
        tmp = dp[state-j][t2[j]]+g[t2[j]][last];
        dp[state][last] = min(dp[state][last], tmp);
    }
}
int main() {
    int n, m, i, j, k, x[20], y[20];
    char s[30];
    for(i = 0; i < 18; i++) t2[1<<i] = i;
    while(scanf("%d %d", &n, &m) == 2) {
        int nut = 1;
        for(i = 0; i < n; i++) {
            scanf("%s", s);
            for(j = 0; s[j]; j++) {
                if(s[j] == '#')
                    x[nut] = i, y[nut++] = j;
                if(s[j] == 'L')
                    x[0] = i, y[0] = j;
            }
        }
        for(i = 0; i < nut; i++) {
            for(j = 0; j < nut; j++) {
                for(k = 0; ; k++)
                    if(x[i]+k >= x[j] && x[i]-k <= x[j] &&
                       y[i]+k >= y[j] && y[i]-k <= y[j]) {
                        g[i][j] = k;
                        break;
                    }
            }
        }
        for(i = 1<<nut; i >= 0; i--)
            for(j = 0; j < nut; j++)
                dp[i][j] = oo;
        dp[0][0] = 0;
        tsp((1<<nut)-1, 0);
        printf("%d\n", dp[(1<<nut)-1][0]);
    }
    return 0;
}