HDU 1045 Fire Net 【连通块的压缩二分图匹配】-优快云博客

题目：http://acm.hdu.edu.cn/showproblem.php?pid=1045

Fire Net

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 15715 Accepted Submission(s): 9519

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

题意概括：

给一张 N*N的图，在上面放炮车，要求炮车不能在同一行或者同一列（除非中间有阻碍物），求最多能放多少炮车。

解题思路：

按照行和列，把会冲突的点压缩成一个点，对压缩后的行和列的点进行二分图匹配。

AC code：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #include <vector>
 6 #define INF 0x3f3f3f3f
 7 using namespace std;
 8 const int MAXN = 15;
 9 char str[MAXN][MAXN];
10 int g[MAXN][MAXN];
11 int linker[MAXN];
12 bool used[MAXN];
13 int xx[MAXN][MAXN], yy[MAXN][MAXN];
14 int uN, vN;
15 
16 bool Find(int x)
17 {
18     for(int i = 1; i <= vN; i++){
19         if(!used[i] && g[x][i]){
20             used[i] = true;
21             if(linker[i] == -1 || Find(linker[i])){
22                 linker[i] = x;
23                 return true;
24             }
25         }
26     }
27     return false;
28 }
29 
30 int match()
31 {
32     int ans = 0;
33     memset(linker, -1, sizeof(linker));
34     for(int i = 1; i <= uN; i++){
35         memset(used, 0, sizeof(used));
36         if(Find(i)) ans++;
37     }
38     return ans;
39 }
40 
41 int main()
42 {
43     int k, row, col;
44     while(~scanf("%d", &k) && k){
45         for(int i = 0; i < k; i++){
46             scanf("%s", str[i]);
47         }
48         memset(xx, 0, sizeof(xx));
49         memset(yy, 0, sizeof(yy));
50         memset(g, 0, sizeof(g));
51         row = col = 0;
52         for(int i = 0; i < k; i++){             //压缩连通块
53             for(int j = 0; j < k; j++){
54                 if(str[i][j] == '.'){
55                     if(j == 0 || str[i][j-1] == 'X') row++;
56                     xx[i][j] = row;
57                 }
58 
59                 if(str[j][i] == '.'){
60                     if(j == 0 || str[j-1][i] == 'X') col++;
61                     yy[j][i] = col;
62                 }
63             }
64         }
65         for(int i = 0; i < k; i++){
66                 for(int j = 0; j < k; j++){
67                     if(str[i][j] == '.')
68                         g[xx[i][j]][yy[i][j]] = 1;
69             }
70         }
71         vN = col, uN = row;
72         printf("%d\n", match());
73     }
74     return 0;
75 }