luogu P2124 奶牛美容

嘟嘟嘟

首先数据范围那么小，那么算法也是相当暴力的。

对于一个点(x, y)所属的联通块，预处理出从这个点出发到这个块外的所有点的曼哈顿距离。复杂度O(n⁴)。

然后求答案：最少答案不一定是三个联通块两两相连，可能是两个联通块之间搭了个桥，然后第三个联通块连接在这个桥上。因此我们像floyd一样，枚举中间点，然后用dis[1][i][j] + dis[2][i][j] + dis[3][i][j]尝试更新答案。

刚开始我因为没有考虑到上面的情况，预处理出每一个点到其他联通块上的点的距离，然后排序，像最小生成树一样贪心取边。导致WA了几个点。

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 #include<cstring>
 #include<cstdlib>
 #include<cctype>
 #include<vector>
 #include<stack>
 #include<queue>
 using namespace std;
 #define enter puts("")
 #define space putchar(' ')
 #define Mem(a, x) memset(a, x, sizeof(a))
 #define rg register
 typedef long long ll;
 typedef double db;
 const int INF = 0x3f3f3f3f;
 const db eps = 1e-;
 const int maxn = ;
 inline ll read()
 {
   ll ans = ;
   char ch = getchar(), last = ' ';
   while(!isdigit(ch)) {last = ch; ch = getchar();}
   while(isdigit(ch)) {ans = ans *  + ch - ''; ch = getchar();}
   if(last == '-') ans = -ans;
   return ans;
 }
 inline void write(ll x)
 {
   if(x < ) x = -x, putchar('-');
   if(x >= ) write(x / );
   putchar(x %  + '');
 }

 int n, m;
 char a[maxn][maxn];
 int vis[maxn][maxn], cnt = ;

 const int dx[] = {-, , , }, dy[] = {, , , -};
 void dfs(int x, int y, int flg)
 {
   vis[x][y] = flg;
   for(int i = ; i < ; ++i)
     {
       int newx = x + dx[i], newy = y + dy[i];
       if(newx >  && newx <= n && newy >  && newy <= m &&
      a[newx][newy] == 'X' && !vis[newx][newy])
     dfs(newx, newy, flg);
     }
 }

 int dis[][maxn][maxn], f[maxn][maxn];
 void solve(int flg, int x, int y)
 {
   for(int i = ; i <= n; ++i)
     for(int j = ; j <= m; ++j)
       dis[flg][i][j] = min(dis[flg][i][j], abs(i - x) + abs(j - y));
 }

 int main()
 {
   n = read(); m = read();
   for(int i = ; i <= n; ++i) scanf("%s", a[i] + );
   for(int i = ; i <= n; ++i)
     for(int j = ; j <= m; ++j)
       if(a[i][j] == 'X' && !vis[i][j])
     dfs(i, j, ++cnt);
   Mem(dis, 0x3f);
   for(int i = ; i <= n; ++i)
     for(int j = ; j <= m; ++j)
       if(a[i][j] == 'X') solve(vis[i][j], i, j);
   Mem(f, 0x3f);
   for(int i = ; i <= n; ++i)
     for(int j = ; j <= m; ++j) if(a[i][j] == 'X')
       {
     f[vis[i][j]][] = min(f[vis[i][j]][], dis[][i][j]);
     f[vis[i][j]][] = min(f[vis[i][j]][], dis[][i][j]);
     f[vis[i][j]][] = min(f[vis[i][j]][], dis[][i][j]);
     f[][vis[i][j]] = f[vis[i][j]][];
     f[][vis[i][j]] = f[vis[i][j]][];
     f[][vis[i][j]] = f[vis[i][j]][];
       }
   int ans = min(f[][] + f[][], min(f[][] + f[][], f[][] + f[][]));
   for(int i = ; i <= n; ++i)
     for(int j = ; j <= m; ++j)
       ans = min(ans, dis[][i][j] + dis[][i][j] + dis[][i][j]);
   write(ans - ); enter;
   return ;
 }