题面
题解
( ext{HNOI2007})真的恐怖
所以这道题非常码
二分答案,将门拆点,于是就变成了一个二分图匹配的题目
反正很恶心
代码
#include<cstdio>
#include<cstring>
#include<cctype>
#include<algorithm>
#include<queue>
#define RG register
#define file(x) freopen(#x".in", "r", stdin);freopen(#x".out", "w", stdout);
#define clear(x, y) memset(x, y, sizeof(x))
inline int read()
{
int data = 0, w = 1; char ch = getchar();
while(ch != '-' && (!isdigit(ch))) ch = getchar();
if(ch == '-') w = -1, ch = getchar();
while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
return data * w;
}
const int N(25), INF(0x3f3f3f3f);
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
int n, m; char g[N][N];
namespace DINIC
{
const int maxn(5e4 + 10), maxm(1e6 + 10);
struct edge { int next, to, cap; } e[maxm];
int head[maxn], e_num = -1, S, T, q[maxn], tail, lev[maxn], cur[maxn];
inline void add_edge(int from, int to, int cap)
{
e[++e_num] = (edge) {head[from], to, cap}; head[from] = e_num;
e[++e_num] = (edge) {head[to], from, 0 }; head[to] = e_num;
}
int bfs()
{
clear(lev, 0); q[tail = lev[S] = 1] = S;
for(RG int i = 1; i <= tail; i++)
{
int x = q[i];
for(RG int j = head[x]; ~j; j = e[j].next)
{
int to = e[j].to;
if(lev[to] || (!e[j].cap)) continue;
lev[q[++tail] = to] = lev[x] + 1;
}
}
return lev[T];
}
int dfs(int x, int f)
{
if(x == T || (!f)) return f;
int ans = 0, cap;
for(RG int &i = cur[x]; ~i; i = e[i].next)
{
int to = e[i].to;
if(e[i].cap && lev[to] == lev[x] + 1)
{
cap = dfs(to, std::min(f - ans, e[i].cap));
e[i].cap -= cap, e[i ^ 1].cap += cap, ans += cap;
if(ans == f) break;
}
}
return ans;
}
inline int main()
{
int ans = 0;
while(bfs())
{
for(RG int i = S; i <= T; i++) cur[i] = head[i];
ans += dfs(S, INF);
}
return ans;
}
}
std::vector<int> dX, dY, pX, pY;
int dis[N][N][N][N];
void BFS(int x, int y, int d[N][N])
{
std::queue<int> qx, qy;
d[x][y] = 0; qx.push(x), qy.push(y);
while(!qx.empty())
{
x = qx.front(); qx.pop();
y = qy.front(); qy.pop();
for(RG int k = 0; k < 4; k++)
{
int tx = x + dx[k], ty = y + dy[k];
if(tx < 1 || tx > n || ty < 1 || ty > m) continue;
if(g[tx][ty] != '.' || d[tx][ty] != -1) continue;
d[tx][ty] = d[x][y] + 1;
qx.push(tx), qy.push(ty);
}
}
}
bool check(int t)
{
int d = dX.size(), p = pX.size(); DINIC::e_num = -1;
clear(DINIC::head, -1); static int idcnt, id_door[N * N][N], id_p[N];
DINIC::S = idcnt = 1;
for(RG int i = 1; i <= t; i++)
for(RG int j = 1; j <= d; j++)
id_door[i][j] = ++idcnt;
for(RG int i = 1; i <= t; i++)
for(RG int j = 1; j <= d; j++)
DINIC::add_edge(DINIC::S, id_door[i][j], 1);
for(RG int i = 1; i <= p; i++)
id_p[i] = ++idcnt;
DINIC::T = ++idcnt;
for(RG int i = 1; i <= p; i++)
DINIC::add_edge(id_p[i], DINIC::T, 1);
for(RG int i = 0; i < d; i++)
for(RG int j = 0; j < p; j++)
{
int ds = dis[dX[i]][dY[i]][pX[j]][pY[j]];
if(ds < 0) continue;
for(RG int k = ds; k <= t; k++)
DINIC::add_edge(id_door[k][i + 1], id_p[j + 1], 1);
}
return DINIC::main() == p;
}
void Solve()
{
int lim = n * m; clear(dis, -1);
for(RG int x = 1; x <= n; x++)
for(RG int y = 1; y <= m; y++)
if(g[x][y] == 'D') dX.push_back(x),
dY.push_back(y), BFS(x, y, dis[x][y]);
else if(g[x][y] == '.') pX.push_back(x), pY.push_back(y);
int L = -1, R = lim + 1, ans = R;
while(L <= R)
{
int mid = (L + R) >> 1;
if(check(mid)) R = mid - 1, ans = mid;
else L = mid + 1;
}
if(ans > lim) puts("impossible");
else printf("%d
", ans);
}
int main()
{
n = read(), m = read();
for(RG int i = 1; i <= n; i++)
scanf("%s", g[i] + 1);
Solve();
return 0;
}