Muddy Fields
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 8434 | Accepted: 3124 |
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
题意：“*”代表沼泽，“。”表示草坪。牛要回家，不能踩草坪，所以只能走沼泽，遇到有沼泽的地方就要主人铺板子，问最少铺多少板子。
思路：典型的二分图匹配中的最小点覆盖。但是不得不说这个建图简直太神奇了！反正我是想不出来，orz。
建图方式按照先行后列。
把行里面连在一起的坑连起来视为一个点，即一块横木板，编上序号
1 0 2 0
0 3 3 3
4 4 4 0
0 0 5 0
再按列做一次则为：
1 0 4 0
0 3 4 5
2 3 4 0
0 0 4 0
一个坑只能被横着的或者被竖着的木板盖住，将原图的坑的也标上不同的序号，一共九个坑
1 . 2 .
. 3 4 5
67 8 .
. . 9 .
比如7号坑可以被横着的4号木板和竖着的3号木板盖住，把每个点的对应的横木板(4)和竖木板(3)中间连一条边的话,则问题转化为 找尽量少的边把这些点都盖住
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
using namespace std;
int n,m;
int cnt1,cnt2;
int G[1010][1010];
int vis[1010];
int link[1010];
char map[110][110];
int m1[110][110];
int m2[110][110];
int dfs(int u)
{
int i;
for(i=0; i<cnt2; i++) {
if(G[u][i]&&!vis[i]) {
vis[i]=1;
if(link[i]==-1|| dfs(link[i])) {
link[i] = u;
return 1;
}
}
}
return 0;
}
void build_tu()
{
int i,j;
cnt1=cnt2=0;
for(i=0; i<n; i++)
for(j=0; j< m; j++) {
if(map[i][j]=='*') {
while(map[i][j]=='*') {
m1[i][j]=cnt1;
j++;
}
cnt1++;
}
}
for(i=0; i<m; i++)
for(j=0; j<n; j++) {
if(map[j][i]=='*') {
while(map[j][i]=='*') {
m2[j][i]=cnt2;
j++;
}
cnt2++;
}
}
/*for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
if(j==m)
printf("%d\n",m1[i][j]);
else
printf("%d ",m1[i][j]);
}
for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
{
if(j==m)
printf("%d\n",m2[i][j]);
else
printf("%d ",m2[i][j]);
}*/
for(i=0; i<n; i++)
for(j=0; j<m; j++) {
if(map[i][j]=='*')
G[m1[i][j]][m2[i][j]]=1;
}
}
int main()
{
int i;
int sum;
while(~scanf("%d %d",&n,&m)) {
sum=0;
memset(G, 0, sizeof(G));
memset(m1, 0, sizeof(m1));
memset(m2, 0, sizeof(m2));
memset(link,-1,sizeof(link));
for(i=0; i<n; i++)
scanf("%s", map[i]);
build_tu();
for(i=0; i<cnt1; i++) {
memset(vis,0,sizeof(vis));
if(dfs(i)) {
//printf("%d---->",dfs(i));
sum++;
}
}
printf("%d\n", sum);
}
return 0;
}