【BZOJ1189】[HNOI2007]紧急疏散evacuate【最大流】【二分】

来源：互联网发布：软件测试职业规划书编辑：程序博客网时间：2024/04/29 11:23

【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;}

0 0