BZOJ1066 [SCOI2007]蜥蜴

首先。。。这是道(很水的)网络流

我们发现"每个时刻不能有两个蜥蜴在同一个柱子上"这个条件是没有用的因为可以让外面的先跳，再让里面的往外跳

但是还有柱子高度的限制，于是把柱子拆点为p1和p2，p1向p2连边，边权为柱子高度

对于相距(注意！是欧几里得距离！)小于d的两个柱子p和q，q2向p1连边，p2向q1连边，边权为inf

S向有蜥蜴的柱子的p1连边，边权为1，可以一步跳出去的柱子p2向T连边，边权为inf

跑最大流即可

 /**************************************************************
     Problem: 1066
     User: rausen
     Language: C++
     Result: Accepted
     Time:136 ms
     Memory:12656 kb
 ****************************************************************/

 #include <cstdio>
 #include <cstring>
 #include <algorithm>

 using namespace std;
 const int N = 1e3 + ;
 const int M = N * N;
 const int inf = 1e9;

 struct edge {
     int next, to, f;
     edge(int _n = , int _t = , int _f = ) : next(_n), to(_t), f(_f) {}
 } e[M];

 int n, m, D, cnt_p, S, T, ans;
 int first[N], tot = ;
 int w[][];
 int d[N], q[N];

 inline void Add_Edges(int x, int y, int f) {
     e[++tot] = edge(first[x], y, f), first[x] = tot;
     e[++tot] = edge(first[y], x, ), first[y] = tot;
 }

 #define y e[x].to
 #define p q[l]
 bool bfs() {
   static int l, r, x;
   memset(d, -, sizeof(d));
   d[q[] = S] = ;
   for (l = r = ; l != r + ; ++l)
     for (x = first[p]; x; x = e[x].next)
       if (!~d[y] && e[x].f) {
     d[q[++r] = y] = d[p] + ;
     if (y == T) return ;
       }
   return ;
 }
 #undef p

 int dfs(int p, int lim) {
   if (p == T || !lim) return lim;
   int x, tmp, rest = lim;
   for (x = first[p]; x && rest; x = e[x].next)
     if (d[y] == d[p] +  && ((tmp = min(e[x].f, rest)) > )) {
       rest -= (tmp = dfs(y, tmp));
       e[x].f -= tmp, e[x ^ ].f += tmp;
       if (!rest) return lim;
     }
   if (rest) d[p] = -;
   return lim - rest;
 }
 #undef y

 inline int Dinic() {
   int res = ;
   while (bfs())
     res += dfs(S, inf);
   return res;
 }

 template <class T> T sqr(T x) {
     return x * x;
 }

 #define in(x, y) w[x][y] * 2 - 1
 #define out(x, y) w[x][y] * 2
 int main() {
     int i, j, k, l;
     char ch;
     scanf("%d%d%d", &n, &m, &D);
     for (cnt_p = , i = ; i <= n; ++i)
         for (j = ; j <= m; ++j) {
             ch = getchar();
             while (ch < '' || ch > '') ch = getchar();
             w[i][j] = ++cnt_p;
             if (ch != '') Add_Edges(in(i, j), out(i, j), ch - '');
         }
     S = cnt_p *  + , T = S + ;
     for (i = ; i <= n; ++i)
         for (j = ; j <= m; ++j)
             for (k = ; k <= n; ++k)
                 for (l = ; l <= m; ++l)
                     if (sqr(i - k) + sqr(j - l) <= D * D && ((i != k) || (j != l)))
                         Add_Edges(out(i, j), in(k, l), inf);
     for (i = ; i <= n; ++i)
         for (j = ; j <= m; ++j) {
             ch = getchar();
             while (ch != '.' && ch != 'L') ch = getchar();
             if (ch == 'L') ++ans, Add_Edges(S, in(i, j), );
             if (i <= D || j <= D || i > n - D || j > m - D)
                 Add_Edges(out(i, j), T, inf);
         }
     printf("%d\n", ans - Dinic());
     return ;
 }
 #undef in
 #undef out