[WC2008]游览计划

标签：math   break   sizeof   cond   get   mem   while   要求   com   

嘟嘟嘟


都说这题是斯坦纳树的板儿题。


斯坦纳树，我也不知道为啥起这么个名儿，斯坦纳树主要用来解决这样一类问题：带边权无向图上有几个（一般约10个）点是【关键点】，要求选择一些边使这些点在同一个联通块内，同时要求所选的边的边权和最小。（摘自兔哥博客）
但说白了就是一种状压dp。令\(dp[i][j][S]\)表示和点\((i, j)\)相连的关键点的状态为\(S\)时的最小代价，于是有两个转移方程：
\[\begin{align*} dp[i][j][S] &= min_{k \in S} \{ dp[i][j][k] + dp[i][j][\complement_kS] - a[i][j] \} \dp[i][j][S] &= min \{ dp[x][y][S] + a[i][j] \} \end{align*}\]
第一个方程很好转移；第二个方程必须满足\((i, j)\)和\((x, y)\)相邻，这个用刷表法就不行了，但是可以用类似spfa的方法转移！

#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 In inline
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 12;
inline ll read()
{
  ll ans = 0;
  char ch = getchar(), last = ' ';
  while(!isdigit(ch)) last = ch, ch = getchar();
  while(isdigit(ch)) ans = (ans << 1) + (ans << 3) + 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, a[maxn][maxn], cnt = 0;
int dp[maxn][maxn][1 << maxn];
struct Node
{
  int x, y, s;
}pre[maxn][maxn][1 << maxn];

#define pr pair<int, int>
#define mp make_pair
queue<pr> q;
bool in[maxn][maxn];
const int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
In void spfa(int s)
{
  while(!q.empty())
    {
      int x = q.front().first, y = q.front().second;
      q.pop(); in[x][y] = 0;
      for(int i = 0, tp; i < 4; ++i)
    {
      int nx = x + dx[i], ny = y + dy[i];
      if(!nx || nx > n || !ny || ny > m) continue;
      if(dp[nx][ny][s] > (tp = dp[x][y][s] + a[nx][ny]))
        {
          dp[nx][ny][s] = tp;
          pre[nx][ny][s] = (Node){x, y, s};       
          if(!in[nx][ny]) q.push(mp(nx, ny)), in[nx][ny] = 1;
        }
    }
    }
}

int vis[maxn][maxn];
In void dfs(int x, int y, int s)
{
  if(!pre[x][y][s].s) return;
  vis[x][y] = 1;
  Node tp = pre[x][y][s];
  dfs(tp.x, tp.y, tp.s);
  if(tp.x == x && tp.y == y) dfs(x, y, s ^ tp.s);
}

int main()
{
  n = read(), m = read();
  Mem(dp, 0x3f);
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j)
      {
    a[i][j] = read(); 
    if(!a[i][j]) dp[i][j][(1 << (cnt++))] = 0;
      }
  for(int s = 0; s < (1 << cnt); ++s)
    {
      for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j)
      {
        for(int k = s, tp; k; k = s & (k - 1))
          if(dp[i][j][s] > (tp = dp[i][j][k] + dp[i][j][s ^ k] - a[i][j]))
        {
          dp[i][j][s] = tp;
          pre[i][j][s] = (Node){i, j, k};
        }
        if(dp[i][j][s] ^ INF) q.push(mp(i, j)), in[i][j] = 1;
      }
      spfa(s);
    }
  int x, y;
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j) if(!a[i][j]) {x = i, y = j; break;}
  write(dp[x][y][(1 << cnt) - 1]), enter;
  dfs(x, y, (1 << cnt) - 1);
  for(int i = 1; i <= n; ++i)
    {
      for(int j = 1; j <= m; ++j)
    if(!a[i][j]) putchar('x');
    else putchar(vis[i][j] ? 'o' : '_');
      enter;
    }
  return 0;
}

[WC2008]游览计划

标签：math   break   sizeof   cond   get   mem   while   要求   com   

原文地址：https://www.cnblogs.com/mrclr/p/10783989.html