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 D: Blackbeard the Pirate

Blackbeard the Pirate has stashed up to 10 treasures on a tropical island, and now he wishes to retrieve them. He is being chased by several authorities however, and so would like to retrieve his treasure as quickly as possible. Blackbeard is no fool; when he hid the treasures, he carefully drew a map of the island which contains the position of each treasure and positions of all obstacles and hostile natives that are present on the island.

Given a map of an island and the point where he comes ashore, help Blackbeard determine the least amount of time necessary for him to collect his treasure.

Input consists of a number of test cases. The first line of each test case contains two integers h and w giving the height and width of the map, respectively, in miles. For simplicity, each map is divided into grid points that are a mile square. The next h lines contain w characters, each describing one square on the map. Each point on the map is one of the following:

•  @  The landing point where Blackbeard comes ashore.
•  ~  Water. Blackbeard cannot travel over water while on the island.
•  #  A large group of palm trees; these are too dense for Blackbeard to travel through.
•  .  Sand, which he can easily travel over.
•  *  A camp of angry natives. Blackbeard must stay at least one square away or risk being captured by them which will terminate his quest. Note, this is one square in any of eight directions, including diagonals.
•  !  A treasure. Blackbeard is a stubborn pirate and will not leave unless he collects all of them.

Blackbeard can only travel in the four cardinal directions; that is, he cannot travel diagonally. Blackbeard travels at a nice slow pace of one mile (or square) per hour, but he sure can dig fast, because digging up a treasure incurs no time penalty whatsoever.

The maximum dimension of the map is 50 by 50. The input ends with a case where both h and w are 0. This case should not be processed.

For each test case, simply print the least number of hours Blackbeard needs to collect all his treasure and return to the landing point. If it is impossible to reach all the treasures, print out -1.

Sample Input

7 7
~~~~~~~
~#!###~
~...#.~
~~....~
~~~.@~~
.~~~~~~
...~~~.
10 10
~~~~~~~~~~
~~!!!###~~
~##...###~
~#....*##~
~#!..**~~~
~~....~~~~
~~~....~~~
~~..~..@~~
~#!.~~~~~~
~~~~~~~~~~
0 0

Output for sample input

10
32


---

Broderick Arneson

題目描述：

海盜黑鬍子要取回寶藏，收集所有寶藏後走回原來的點，盤面上最多 10 個寶藏。

途中不能經過水路，也不能通過叢林，陸地上的原住民會攻擊！

黑鬍子只能走上下左右，但是原住民卻鄰近八個方格攻擊其他敵人。

題目解法：

請小心只有 1 個點的 TSP。

當然有可能寶藏在原住民的攻擊範圍內，因此答案一定是 impossible。

加上出發點總共是 11 個點的 TSP，請注意記憶體。

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
using namespace std;
char g[55][55], cannot[55][55];
int ng[55][55], used[55][55], h, w;
int dp[1<<12][12];
int dist[12][12];
void bfs(int x, int y) {
    queue<int> X, Y;
    int i, j, tx, ty;
    int dx[] = {0,0,1,-1};
    int dy[] = {1,-1,0,0};
    X.push(x), Y.push(y);
    memset(used, 0, sizeof(used));
    int st = ng[x][y];
    if(cannot[x][y])    return;
    while(!X.empty()) {
        x = X.front(), X.pop();
        y = Y.front(), Y.pop();
        for(i = 0; i < 4; i++) {
            tx = x+dx[i], ty = y+dy[i];
            if(tx < 0 || ty < 0 || tx >= h || ty >= w)
                continue;
            if(cannot[tx][ty])    continue;
            if(g[tx][ty] == '!' || g[tx][ty] == '.') {
                if(used[tx][ty] == 0) {
                    used[tx][ty] = used[x][y]+1;
                    X.push(tx), Y.push(ty);
                    if(g[tx][ty] == '!')
                        dist[st][ng[tx][ty]] = used[x][y]+1;
                }
            }
        }
    }   
}
int dfs(int state, int last, int n) {
    int &ret = dp[state][last];
    if(ret != -1)    return ret;
    if(state == 0)
        return dist[last][0];
    ret = 0xfffffff;
    int i;
    for(i = 0; i < n; i++) {
        if((state>>i)&1) {
            ret = min(ret, dfs(state^(1<<i), i, n)+dist[last][i]);
        }
    }
    return ret;
}
int main() {
    int i, j, k;
    while(scanf("%d %d", &h, &w) == 2 && h+w) {
        for(i = 0; i < h; i++)
            scanf("%s", g[i]);
        memset(dp, -1, sizeof(dp));
        memset(cannot, 0, sizeof(cannot));
        int n = 0;
        int dx[] = {0,0,1,1,1,-1,-1,-1};
        int dy[] = {1,-1,0,1,-1,0,1,-1};
        int tx, ty;
        for(i = 0; i < h; i++) {
            for(j = 0; j < w; j++) {
                if(g[i][j] == '@')    g[i][j] = '!';
                if(g[i][j] == '!')
                    ng[i][j] = n++;
                if(g[i][j] == '*') {
                    for(k = 0; k < 8; k++) {
                        tx = i+dx[k], ty = j+dy[k];
                        if(tx < 0 || ty < 0 || tx >= h || ty >= w)
                            continue;
                        cannot[tx][ty] = 1;
                    }
                }
                if(g[i][j] == '~')    cannot[i][j] = 1;
                if(g[i][j] == '#')    cannot[i][j] = 1;
            }
        }
        for(i = 0; i < n; i++) {
            for(j = 0; j < n; j++)
                dist[i][j] = 0xfffffff;
        }
        for(i = 0; i < h; i++) {
            for(j = 0; j < w; j++) {
                if(g[i][j] == '!')
                    bfs(i, j);
            }
        }
        int ret = dfs((1<<n)-1-1, 0, n);
        if(ret == 0xfffffff)    ret = -1;
        if(n == 1)                ret = 0;
        printf("%d\n", ret);
    }
    return 0;
}