Muddy Fields
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8881 | Accepted: 3300 |
Description
Rain has pummeled the cows‘ field, a rectangular grid of R rows and C columns (1 <= R <= 50, 1 <= C <= 50). While good for the grass, the rain makes some patches of bare earth quite muddy. The cows, being meticulous grazers, don‘t want to get their hooves dirty while they eat.
To prevent those muddy hooves, Farmer John will place a number of wooden boards over the muddy parts of the cows‘ field. Each of the boards is 1 unit wide, and can be any length long. Each board must be aligned parallel to one of the sides of the field.
Farmer John wishes to minimize the number of boards needed to cover the muddy spots, some of which might require more than one board to cover. The boards may not cover any grass and deprive the cows of grazing area but they can overlap each other.
Compute the minimum number of boards FJ requires to cover all the mud in the field.
Input
* Line 1: Two space-separated integers: R and C
* Lines 2..R+1: Each line contains a string of C characters, with ‘*‘ representing a muddy patch, and ‘.‘ representing a grassy patch. No spaces are present.
Output
* Line 1: A single integer representing the number of boards FJ needs.
Sample Input
4 4 *.*. .*** ***. ..*.
Sample Output
4
Hint
OUTPUT DETAILS:
Boards 1, 2, 3 and 4 are placed as follows:
1.2.
.333
444.
..2.
Board 2 overlaps boards 3 and 4.
Source
题目意思：
一个n*m的图，其中‘.‘表示平地，‘*‘水池。现用一些宽为1个单位，长度不限的木板覆盖所有的水池（某个水池可以被覆盖多次）。问所用木板最少为多少。
思路：
每个水池可以被横着覆盖也可以被竖着覆盖，那么题目就成为选用一些覆盖策略使得所有点被横着覆盖或者被竖着覆盖，而某些点被横着或竖着覆盖可以影响与其横着连续的点或竖着连续的点。那么横着处理一下图如同题目中Hint中，同理竖着处理一下图。每个点有两个数字，一个是横着状态数字，另一个是竖着状态数字，连边后建图，就是最小点覆盖了。
代码：
1 #include <cstdio>
2 #include <cstring>
3 #include <algorithm>
4 #include <iostream>
5 #include <vector>
6 #include <queue>
7 #include <cmath>
8 #include <set>
9 using namespace std;
10
11 #define N 50
12
13 int max(int x,int y){return x>y?x:y;}
14 int min(int x,int y){return x<y?x:y;}
15 int abs(int x,int y){return x<0?-x:x;}
16
17 int n, m;
18 vector<int>ve[N*N];
19 int from[N*N];
20 bool visited[N*N];
21 char map[N][N];
22 int map1[N][N];
23 int map2[N][N];
24
25 int march(int u){
26 int i, v;
27 for(i=0;i<ve[u].size();i++){
28 v=ve[u][i];
29 if(!visited[v]){
30 visited[v]=true;
31 if(from[v]==-1||march(from[v])){
32 from[v]=u;
33 return 1;
34 }
35 }
36 }
37 return 0;
38 }
39
40 main()
41 {
42 int i, j, k;
43 while(scanf("%d %d",&n,&m)==2){
44 for(i=0;i<n;i++) scanf("%s",map[i]);
45 memset(map1,-1,sizeof(map1));
46 memset(map2,-1,sizeof(map2));
47 int num=1;
48 int maxh=0;
49 //横着处理
50 for(i=0;i<n;i++){
51 for(j=0;j<m;j++){
52 if(map[i][j]==‘*‘){
53 if(j==0) map1[i][j]=num;
54 else{
55 if(map[i][j-1]==‘*‘) map1[i][j]=map1[i][j-1];
56 else map1[i][j]=num;
57 }
58 }
59 else {
60 if(j<m-1&&map[i][j+1]==‘*‘) num++;
61 }
62 }
63 num++;
64 maxh=max(maxh,num);
65 }
66 //竖着处理
67 num=1;
68 for(j=0;j<m;j++){
69 for(i=0;i<n;i++){
70 if(map[i][j]==‘*‘){
71 if(i==0) map2[i][j]=num;
72 else{
73 if(map[i-1][j]==‘*‘) map2[i][j]=map2[i-1][j];
74 else map2[i][j]=num;
75 }
76 }
77 else{
78 if(i<n-1&&map[i+1][j]==‘*‘) num++;
79 }
80 }
81 num++;
82 maxh=max(maxh,num);
83 }
84 //建二分图
85 for(i=0;i<maxh;i++) ve[i].clear();
86 for(i=0;i<n;i++){
87 for(j=0;j<m;j++){
88 if(map1[i][j]!=-1&&map2[i][j]!=-1){
89 ve[map1[i][j]].push_back(map2[i][j]);
90 }
91 }
92 }
93 //二分匹配
94 memset(from,-1,sizeof(from));
95 num=0;
96 for(i=0;i<maxh;i++){
97 memset(visited,false,sizeof(visited));
98 if(march(i)) num++;
99 }
100 printf("%d\n",num);
101 }
102 }