Fire Net
Time Limit: 2000/1000 MS (java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 11193 Accepted Submission(s): 6673
PRoblem Description Suppose that we have a square city with straight streets. A map of a city is a square board with n rows and n columns, each representing a street or a piece of wall.
A blockhouse is a small castle that has four openings through which to shoot. The four openings are facing North, East, South, and West, respectively. There will be one machine gun shooting through each opening.
Here we assume that a bullet is so powerful that it can run across any distance and destroy a blockhouse on its way. On the other hand, a wall is so strongly built that can stop the bullets.
The goal is to place as many blockhouses in a city as possible so that no two can destroy each other. A configuration of blockhouses is legal provided that no two blockhouses are on the same horizontal row or vertical column in a map unless there is at least one wall separating them. In this problem we will consider small square cities (at most 4x4) that contain walls through which bullets cannot run through.
The following image shows five pictures of the same board. The first picture is the empty board, the second and third pictures show legal configurations, and the fourth and fifth pictures show illegal configurations. For this board, the maximum number of blockhouses in a legal configuration is 5; the second picture shows one way to do it, but there are several other ways.
Your task is to write a program that, given a description of a map, calculates the maximum number of blockhouses that can be placed in the city in a legal configuration.
Input The input file contains one or more map descriptions, followed by a line containing the number 0 that signals the end of the file. Each map description begins with a line containing a positive integer n that is the size of the city; n will be at most 4. The next n lines each describe one row of the map, with a ‘.’ indicating an open space and an uppercase ‘X’ indicating a wall. There are no spaces in the input file.
Output For each test case, output one line containing the maximum number of blockhouses that can be placed in the city in a legal configuration.
Sample Input 4 .X.. …. XX.. …. 2 XX .X 3 .X. X.X .X. 3 … .XX .XX 4 …. …. …. …. 0
Sample Output 5 1 5 2 4
題意：炮臺不能同行同列存在，除非中間隔著墻 n<=4,直接dfs即可
#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <map>#define MST(s,q) memset(s,q,sizeof(s))#define INF 0x3f3f3f3f#define MAXN 1005using namespace std;char Map[4][4];struct node{ int x, y;} Node[20];int num , ans, n; // num指能放炮臺的位置的數量int move_x[4] = {0, 0, 1, -1}, move_y[4] = {1, -1, 0, 0};bool CanPut(int k) // 第k個點能不能放{ for (int i = 0; i < 4; i++) // 向四個方向遍歷 { int x = Node[k].x + move_x[i]; int y = Node[k].y + move_y[i]; while (x >= 0 && x < n && y >= 0 && y < n) { if (Map[x][y] == 'X') break; else if (Map[x][y] == '#') // 遇到其他炮臺，則不能放 return 0; x += move_x[i]; y += move_y[i]; } } return 1;}void dfs(int x, int V) // 從第0個點開始dfs這num個點{ if (V > ans) ans = V; if (x == num) return; dfs(x + 1, V); // x位置不放炮臺 if (CanPut(x)) // x位置放炮臺 { Map[Node[x].x][Node[x].y] = '#'; dfs(x + 1, V + 1); Map[Node[x].x][Node[x].y] = '.'; }}int main(){ while (cin >> n, n) { for (int i = 0; i < n; i++) scanf("%s", Map[i]); num = 0; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) { if (Map[i][j] == '.') { Node[num].x = i, Node[num].y = j; num++; } } ans = 0; dfs(0, 0); printf("%d/n", ans); }}新聞熱點
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