Fire Net

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8620    Accepted Submission(s):
4976

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

Source

Zhejiang
University Local Contest 2001

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二分图行列匹配， 主要是处理过程；

#include <cstdio>
#include <cstring>
using namespace std;
char Ap[5][5];
int Gra[5][5], vis[5], dis[5];
int dx[5], dy[5], left[5], right[5];
int n;
struct Node{
    int x, y;
}a[5][5];
void Printf(){
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            printf("%c", Ap[i][j]);
        }
        printf("\n");
    }
}
int x, y;
bool Search(int a){
    for(int i = 1; i <= y; i++){
        if(Gra[a][i] && !vis[i]){
            vis[i] = 1;
            if(!dis[i] || Search(dis[i])){
                dis[i] = a;
                return true;
            }
        }
    }
    return false;
}
int main(){
    while(scanf("%d", &n) != EOF && n){
        for(int i = 0; i < n; i++)
        {
            getchar();
            for(int j = 0; j < n; j++)
                scanf("%c", &Ap[i][j]);
        }
        x = y = 0;
        for(int i = 0; i < n; i++)   //行列匹配；
            for(int j = 0; j < n; j++)
            {
                if(Ap[i][j] == ‘.‘)
                {
                    if(j == 0 || Ap[i][j-1] == ‘X‘){
                        x++;        /***********************/
                    }
                    a[i][j].x = x;
                }
                if(Ap[j][i] == ‘.‘)
                {
                    if(j == 0 || Ap[j-1][i] == ‘X‘){
                        y++;       /***********************/
                    }
                    a[j][i].y = y;
                }
            }
        memset(Gra, 0, sizeof(Gra));
        for(int i = 0; i < n; i++)
            for(int j = 0; j < n; j++){
                if(Ap[i][j] == ‘.‘){
                    int u = a[i][j].x;
                    int v = a[i][j].y;
                    Gra[u][v] = 1;
                }
            }
        int ans = 0;
        memset(dis, 0, sizeof(dis));
        for(int i = 1; i <= x; i++)
        {
            memset(vis, 0, sizeof(vis));
            if(Search(i))
                ans++;
        }
        printf("%d\n", ans);
    }
    return 0;
} 

暴搜；

#include <cstdio>
int n, cnt;
char map[5][5];
bool Build(int row, int col){
    int i, j;
    for(int i = row; i >= 0; i--){
        if(map[i][col] == ‘O‘)
            return false;
        if(map[i][col] == ‘X‘)
            break;
    }
    for(int i = col; i >= 0; i--){
        if(map[row][i] == ‘O‘)
            return false;
        if(map[row][i] == ‘X‘)
            break;
    }
    return true ;
}
void Dfs(int i, int num){
    if(i == n*n){
    //    printf("%d %d\n", i, num);
        if(num > cnt)
            cnt = num;
        return;
    }
    else{
        int row = i / n;
        int col = i % n;
        if(map[row][col] == ‘.‘ && Build(row, col)){
            map[row][col] = ‘O‘;
            Dfs(i+1, num+1);
            map[row][col] = ‘.‘;
        }
        Dfs(i+1, num);
    }
}
int main(){
    while(scanf("%d", &n), n){
        cnt = 0;
        for(int i = 0; i < n; i++)
            scanf("%s", map[i]);
        Dfs(0, 0);
        printf("%d\n", cnt);
    }
    return 0;
}