[ZOJ 1002] Fire Net

[ruiming]

ZOJ1002 - Fire Net

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.

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.

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

分析

这道题的意思很明显，在一张4X4的地图中放置碉堡，碉堡不能同行或同列，除非中间有墙阻隔。问最多能放置多少个碉堡。

做这道题的时候还不了解二分图，所以用的DFS方法。此题最佳解法应是使用二分图。

关于DFS的做法这里不再多说，以后再改用二分图实现，这里简单说下二分图的做法。

以下面这个图为例子

.X..
....
XX..
....

从左到右分别用1 2 3 ...表示，墙壁不编号也不打断编号。

横向可以得到多条边：

(1) (2 3) (4 5 6 7) (8 9) (10 11 12 13)

纵向可以得到多条边：

(1 4) (5) (2 6 8 12) (3 7 9 13) (10) (11)

然后我们可以将有横向和纵向有覆盖的边连接起来。接着就是找最大匹配的问题。

匈牙利算法是从出发点开始遍历，对于每一个出发点，广度优先找到一个终点，如果该点没有被访问，置该点为被访问，否则继续广度搜索下去。对于这个终点，判断其是否已经被匹配，如果被匹配了，将其匹配的起始点递归调用自身，否则，建立起点到终点的匹配，返回成功。

递归调用自身的时候，该起点会尝试找到下一个可以匹配的点，找到后修改匹配返回成功。如果找不到则返回失败。

这个递归调用的过程要搞清楚。需要两个数组一个保存匹配，一个在递归调用的时候保存已被访问的节点，但是这个数组在外层需要不断清空。

程序

#include <iostream>
#include <cstring>
using namespace std;
int n;
int row, col;
int number = 0;
int result = 0;
int G[4][4];
int len[32][4];
int mat[64], used[64];
struct Edge {
    int start, end;
}e[100];

// 找下一个匹配
bool find(int start) {
    for (int i = 0; i < number; i++) {
        if (e[i].start == start) {
            int end = e[i].end;
            // 终点没有被占用
            if (!used[end]) {
                // 标记占用
                used[end] = 1;
                // 终点没有被匹配或者终点对应的起始点能找到另一个匹配
                if (mat[end] == -1 || find(mat[end])) {
                    // 终点匹配起始点
                    mat[end] = start;
                    return true;
                }
            }
        }
    }
    return false;
}

// 匈牙利算法
void hungary() {
    memset(mat, -1, sizeof(mat));
    memset(used, 0, sizeof(used));
    for (int i = 0; i < row; i++) {
        if (find(i)) {
            result++;
        }
        memset(used, 0, sizeof(used));
    }
}

void build() {
    for (int i = 0; i < row; i++) {
        for (int j = row; j < col; j++) {
            int k = 0;
            bool connect = false;
            while (len[i][k] != 0 && k<4) {
                int u = 0;
                while (len[j][u] != 0 && u<4) {
                    if (len[j][u] == len[i][k]) {
                        Edge a;
                        a.start = i;
                        a.end = j;
                        e[number++] = a;
                        connect = true;
                        break;
                    }
                    u++;
                }
                k++;
                if (connect) {
                    break;
                }
            }
        }
    }
}

int main() {
    while (cin >> n && n) {
        int k = 0, num = 1;
        number = 0, result = 0;
        memset(G, 0, sizeof(G));
        memset(len, 0, sizeof(len));
        char temp;
        // 编号, 数组 G
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                cin >> temp;
                if (temp != 'X') {
                    G[i][j] = num;
                    num++;
                } else {
                    G[i][j] = 0;
                }
            }
        }
        // 建立横向边数组 len
        for (int i = 0; i < n; i++) {
            int u = 0;
            bool stone = true;
            for (int j = 0; j < n; j++) {
                if (G[i][j] != 0) {
                    stone = false;
                    len[k][u] = G[i][j];
                    u++;
                } else {
                    if (stone)    continue;
                    else {
                        stone = true;
                        k++;
                        u = 0;
                    }
                }
            }
            if (!stone)    k++;
        }
        row = k;
        // 建立纵向边数组 len 接上, 以 row 为分割
        for (int j = 0; j < n; j++) {
            int u = 0;
            bool stone = true;
            for (int i = 0; i < n; i++) {
                if (G[i][j] != 0) {
                    stone = false;
                    len[k][u] = G[i][j];
                    u++;
                } else {
                    if (stone)    continue;
                    else {
                        stone = true;
                        k++;
                        u = 0;
                    }
                }
            }
            if (!stone)    k++;
        }
        col = k;
        // 建立二分图
        build();
        // 找最大匹配
        hungary();
        cout << result << endl;
    }
    return 0;
}