USACO 2011 Nov. [Bronze] P4. Cow Beauty Pageant

原题地址：http://www.usaco.org/index.php?page=viewproblem2&cpid=87

题目描述
Hearing that the latest fashion trend was cows with two spots on their hides, Farmer John has purchased an entire herd of two-spot cows.
Unfortunately, fashion trends tend to change quickly, and the most popular current fashion is cows with only one spot!
FJ wants to make his herd more fashionable by painting each of his cows in such a way that merges their two spots into one. The hide of a cow is represented by an N by M (1 <= N,M <= 50) grid of characters like this:

................
..XXXX....XXX...
...XXXX....XX...
.XXXX......XXX..
........XXXXX...
.........XXX....

Here, each ‘X’ denotes part of a spot. Two 'X’s belong to the same spot if they are vertically or horizontally adjacent (diagonally adjacent does not count), so the figure above has exactly two spots. All of the cows in FJ’s herd have exactly two spots.
FJ wants to use as little paint as possible to merge the two spots into one. In the example above, he can do this by painting only three
additional characters with 'X’s (the new characters are marked with '*'s below to make them easier to see).

................
..XXXX....XXX...
...XXXX*...XX...
.XXXX..**..XXX..
........XXXXX...
.........XXX....

Please help FJ determine the minimum number of new 'X’s he must paint in order to merge two spots into one large spot.

听说最近的流行趋势是牛皮上有两个斑点的奶牛，农场主约翰买下了整群的有两个斑点的奶牛。
不幸的是，流行趋势往往会迅速发生变化，目前最流行的时尚是只有一个斑点的奶牛！
FJ想用画的方法让他的每一头奶牛的两个斑点连在一起成为一个斑点，从而使他的牛群更时尚。奶牛的皮由一个n*m（1<=n，m<＝50）的字符格子表示：

................
..XXXX....XXX...
...XXXX....XX...
.XXXX......XXX..
........XXXXX...
.........XXX....

这里，每一个’X’表示一个斑点的一部分。垂直或水平相邻（对角线相邻不在内）的两个’X’属于同一个斑点，所以上图正好有两个斑点。所有FJ的奶牛正好都只有两个斑点。
FJ希望使用尽可能少的油漆合并两个斑点为一个。在上面的例子中，他只要画三个附加的字符就可以做到这一点。
附加字符“X”（新的字符如下图中的*那样表示，很容易看到）。

................
..XXXX....XXX...
...XXXX*...XX...
.XXXX..**..XXX..
........XXXXX...
.........XXX....

请帮助FJ确定最少的画的点的数目从而能让两个斑点合并为一个。

输入
* Line 1: Two space-separated integers, N and M.
* Lines 2…1+N: Each line contains a length-M string of 'X’s and '.'s specifying one row of the cow hide pattern.

第1行：两个空格隔开的整数，N和M。
第2…1+N行：每行包含一个长度为M的字符串，只有’X’和’.'组成，表示一只奶牛的兽皮图案。

输出
* Line 1: The minimum number of new 'X’s that must be added to the input pattern in order to obtain one single spot.

第1行：最小的添加新的’X’字符的数目从而让两块斑点能合并为单一的一块斑点。

样例
输入

6 16
................
..XXXX....XXX...
...XXXX....XX...
.XXXX......XXX..
........XXXXX...
.........XXX....

输出

提示
INPUT DETAILS:
The pattern in the input shows a cow hide with two distinct spots, labeled 1 and 2 below:

................
..1111....222...
...1111....22...
.1111......222..
........22222...
.........222....

OUTPUT DETAILS:
Three 'X’s suffice to join the two spots into one:

................
..1111....222...
...1111X...22...
.1111..XX..222..
........22222...
.........222....

输入描述：
输入的奶牛兽皮图案有两块明显的斑点，用标签1和标签2分别如下表示：

................
..1111....222...
...1111....22...
.1111......222..
........22222...
.........222....

输出描述：
用三个’X’足够让两块斑点变成一块。

................
..1111....222...
...1111X...22...
.1111..XX..222..
........22222...
.........222....

题意：
一张图中有2个连通块，求把这2个连通块连到一起最少需要填充多少个点。

思路：
用dfs遍历其中一个连通块，打上标记来区分两个连通块。
根据标记分别存放两个连通块。
求分别属于两个连通块的点之间的曼哈顿距离（暴力循环），取最小值。

AC 代码：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
#define map Mapint n,m;
int map[57][57];
int dx[4]={1,0,-1,0},dy[4]={0,1,0,-1};void init(){//存图char c;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>c;if(c=='X')map[i][j]=1;}}
}void dfs(int x,int y){//遍历1个连通块map[x][y]=2;//打标记for(int i=0;i<4;i++){if(map[ x+dx[i] ][ y+dy[i] ]==1){dfs(x+dx[i],y+dy[i]);}}
}struct node{int x,y;
};node pot1[2500],pot2[2500];//两个连通块
int p1,p2;void savepot(){//分别存两个连通块for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(map[i][j]==2){pot1[p1]={i,j};p1++;}else if(map[i][j]==1){pot2[p2]={i,j};p2++;}}}
}int mht(){//求两组点之间的曼哈顿距离，取最小值int min0=101,mht0;for(int i=0;i<p1;i++){for(int j=0;j<p2;j++){mht0=abs(pot1[i].x-pot2[j].x)+abs(pot1[i].y-pot2[j].y)-1;min0=min(mht0,min0);}}return min0;
}node first;
void findfirst(){//任意找一个点作为dfs的起始点for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(map[i][j]==1){first={i,j};return;}}}
}int main(){cin>>n>>m;init();findfirst();dfs(first.x,first.y);savepot();int ans=mht();cout<<ans;return 0;
}

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