[国家集训队]部落战争

嘟嘟嘟


这其实就是一道最小边覆盖的板儿题。


暴力连边，然后跑匈牙利（dinic），则答案就是总结点数-匹配数。
比如节点1和2,2和3匹配上了，那么就是1到2,2到3这两条路径连接到一块，相当于把节点3合并到这条路径上了。所以从路径数就是总结点数-连接次数。


然而我不知怎么想的写了个费用流，虽然正确性是对的，但是因为是一条条增广路找，就慢了很多……

#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 = 55;
const int maxN = 5e3 + 5;
const int maxe = 1e7 + 5;
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');
}

char s[maxn][maxn];
int n, m, r, c, t;
struct Edge
{
  int nxt, from, to, cap, cos;
}e[maxe];
int head[maxN], ecnt = -1;
In void addEdge(int x, int y, int w, int c)
{
  e[++ecnt] = (Edge){head[x], x, y, w, c};
  head[x] = ecnt;
  e[++ecnt] = (Edge){head[y], y, x, 0, -c};
  head[y] = ecnt;
}

bool in[maxN];
int dis[maxN], pre[maxN], flow[maxN];
In bool spfa()
{
  Mem(dis, 0x3f), Mem(in, 0);
  dis[0] = 0, flow[0] = INF;
  queue<int> q; q.push(0);
  while(!q.empty())
    {
      int now = q.front(); q.pop(); in[now] = 0;
      for(int i = head[now], v; ~i; i = e[i].nxt)
	{
	  if(dis[v = e[i].to] > dis[now] + e[i].cos && e[i].cap > 0)
	    {
	      dis[v] = dis[now] + e[i].cos;
	      pre[v] = i;
	      flow[v] = min(flow[now], e[i].cap);
	      if(!in[v]) q.push(v), in[v] = 1;
	    }
	}
    }
  return dis[t] ^ INF;
}
int minCost = 0;
In void update()
{
  int x = t;
  while(x)
    {
      int i = pre[x];
      e[i].cap -= flow[t];
      e[i ^ 1].cap += flow[t];
      x = e[i].from;
    }
  minCost += flow[t] * dis[t];
}

In int MCMF()
{
  while(spfa()) update();
  return minCost;
}

int dx[5], dy[5], DIR;
In void init()
{
  if(r == c)
    {
      DIR = 2;
      dx[0] = r, dx[1] = r;
      dy[0] = r, dy[1] = -r;
    }
  else
    {
      DIR = 4;
      dx[0] = r, dx[1] = c, dx[2] = c, dx[3] = r;
      dy[0] = c, dy[1] = r, dy[2] = -r, dy[3] = -c;
    }
}

In int num(int x, int y, int z) {return (x - 1) * m + y + z * n * m;}
In void build(int x, int y)
{
  addEdge(0, num(x, y, 0), 1, 1);
  addEdge(num(x, y, 1), t, 1, 0);
  for(int i = 0; i < DIR; ++i)
    {
      int nx = x + dx[i], ny = y + dy[i];
      if(nx > 0 && nx <= n && ny > 0 && ny <= m && s[nx][ny] == '.')
	addEdge(num(x, y, 0), num(nx, ny, 1), 1, 0);
    }
}

int main()
{
  //freopen("ha.in", "r", stdin);
  Mem(head, -1);
  n = read(), m = read(), r = read(), c = read();
  t = n * m * 2 + 1;
  for(int i = 1; i <= n; ++i) scanf("%s", s[i] + 1);
  init(); int tot = 0;
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j) if(s[i][j] == '.') build(i, j), ++tot;
  write(tot - MCMF()), enter;
  return 0;
}