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-8;
const int maxn = 55;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) {last = ch; ch = getchar();}
  while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
  if(last == '-') ans = -ans;
  return ans;
}
inline void write(ll x)
{
  if(x < 0) x = -x, putchar('-');
  if(x >= 10) write(x / 10);
  putchar(x % 10 + '0');
}

int n, m;
char a[maxn][maxn];
int vis[maxn][maxn], cnt = 0;

const int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
void dfs(int x, int y, int flg)
{
  vis[x][y] = flg;
  for(int i = 0; i < 4; ++i)
    {
      int newx = x + dx[i], newy = y + dy[i];
      if(newx > 0 && newx <= n && newy > 0 && newy <= m &&
     a[newx][newy] == 'X' && !vis[newx][newy])
    dfs(newx, newy, flg);
    }
}

int dis[4][maxn][maxn], f[maxn][maxn];
void solve(int flg, int x, int y)
{
  for(int i = 1; i <= n; ++i)
    for(int j = 1; 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 = 1; i <= n; ++i) scanf("%s", a[i] + 1);
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j)
      if(a[i][j] == 'X' && !vis[i][j])
    dfs(i, j, ++cnt);
  Mem(dis, 0x3f);
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j)
      if(a[i][j] == 'X') solve(vis[i][j], i, j);
  Mem(f, 0x3f);
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j) if(a[i][j] == 'X')
      {
    f[vis[i][j]][1] = min(f[vis[i][j]][1], dis[1][i][j]);
    f[vis[i][j]][2] = min(f[vis[i][j]][2], dis[2][i][j]);
    f[vis[i][j]][3] = min(f[vis[i][j]][3], dis[3][i][j]);
    f[1][vis[i][j]] = f[vis[i][j]][1];
    f[2][vis[i][j]] = f[vis[i][j]][2];
    f[3][vis[i][j]] = f[vis[i][j]][3];
      }
  int ans = min(f[1][2] + f[2][3], min(f[1][3] + f[1][2], f[1][3] + f[2][3]));
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j)
      ans = min(ans, dis[1][i][j] + dis[2][i][j] + dis[3][i][j]);
  write(ans - 2); enter;
  return 0;
}