【题目链接】
先处理出每个人到每个门的最短距离。
二分答案mid。
从S到每个人连边,容量为1。
每个门拆为mid个点,第i个点代表第i个时刻是否的门,每个人向 第(最短距离)个点 ~ 第mid个点 连边。
每个门向T连边。
跑最大流,看是否满流。
一直担心建图错,结果没错,倒是忘了判impossible了。
/* Pigonometry */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int maxn = 50005, maxm = 400005, maxg = 22, maxq = 1000000, inf = 0x3f3f3f3f; int n, m, k, tot, peo, mp[maxg][maxg], head[maxn], cur[maxn], cnt, q[maxq]; int bg, ed, dis[maxg][maxg][maxg << 2], id[maxg][maxg], door[maxg << 2][402], depth[maxn]; int dx[] = {1, 0, -1, 0}, dy[] = {0, -1, 0, 1}; struct _edge { int v, w, next; } g[maxm << 1]; struct _data { int x, y, s; } q2[maxq]; inline void getdis(int x, int y, int no) { int h = 0, t = 0; dis[x][y][no] = 0; q2[t++] = (_data){x, y, 0}; while(h != t) { _data u = q2[h++]; for(int i = 0; i < 4; i++) { int xx = u.x + dx[i], yy = u.y + dy[i]; if(mp[xx][yy] != 1) continue; if(dis[xx][yy][no] > u.s + 1) { dis[xx][yy][no] = u.s + 1; q2[t++] = (_data){xx, yy, u.s + 1}; } } } } inline void add(int u, int v, int w) { g[cnt] = (_edge){v, w, head[u]}; head[u] = cnt++; } inline void insert(int u, int v, int w) { add(u, v, w); add(v, u, 0); } inline bool bfs() { for(int i = 0; i <= tot; i++) depth[i] = -1; int h = 0, t = 0, u, i; depth[q[t++] = bg] = 1; while(h != t) for(i = head[u = q[h++]]; ~i; i = g[i].next) if(g[i].w && !~depth[g[i].v]) { depth[g[i].v] = depth[u] + 1; if(g[i].v == ed) return 1; q[t++] = g[i].v; } return 0; } inline int dfs(int x, int flow) { if(x == ed) return flow; int left = flow; for(int i = cur[x]; ~i; i = g[i].next) if(g[i].w && depth[g[i].v] == depth[x] + 1) { int tmp = dfs(g[i].v, min(left, g[i].w)); left -= tmp; g[i].w -= tmp; g[i ^ 1].w += tmp; if(g[i].w) cur[x] = i; if(!left) return flow; } if(left == flow) depth[x] = -1; return flow - left; } inline bool check(int mid) { for(int i = 0; i <= tot; i++) head[i] = -1; cnt = 0; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(mp[i][j] == 1) insert(bg, id[i][j], 1); for(int i = 1; i <= k; i++) for(int j = 1; j <= mid; j++) insert(door[i][j], ed, 1); for(int t = 1; t <= k; t++) for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(mp[i][j] == 1 && dis[i][j][t] <= mid) for(int l = dis[i][j][t]; l <= mid; l++) insert(id[i][j], door[t][l], 1); int ans = peo; while(bfs()) { for(int i = 0; i <= tot; i++) cur[i] = head[i]; ans -= dfs(bg, inf); } return !ans; } char str[maxg]; int main() { scanf("%d%d", &n, &m); k = peo = tot = 0; for(int i = 1; i <= n; i++) { scanf("%s", str + 1); for(int j = 1; j <= m; j++) if(str[j] == '.') mp[i][j] = 1, peo++; else if(str[j] == 'D') mp[i][j] = 2; } memset(dis, 0x3f, sizeof(dis)); bg = ++tot; ed = ++tot; for(int i = 1; i <= n; i++) for(int j = 1; j <= m; j++) if(mp[i][j] == 1) id[i][j] = ++tot; else if(mp[i][j] == 2) { getdis(i, j, ++k); for(int t = 1; t <= 400; t++) door[k][t] = ++tot; } int l = 0, r = 400, ans = inf; while(l <= r) { int mid = l + r >> 1; if(check(mid)) ans = mid, r = mid - 1; else l = mid + 1; } if(ans == inf) printf("impossible\n"); else printf("%d\n", ans); return 0; }