hdu3681 Prison Break (dfs+二分+状态压缩)

题意：一个机器人一开始在F，D表示障碍，Y表示开关，G表示充电器。

机器人每走一步消耗一点能量，走到充电器可以瞬间充满电（每个充电器只能充一次），现在要打开所有的开关（每个开关也只用打开一次），问最少的初始电量。

首先计算出所有的G和Y两两之间的最短路，然后二分答案，用dfs判断是否能走完。

代码里有注释：

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <queue>#include <vector>#define INF 0x3f3f3f3fusing namespace std;struct Point{int x, y;Point() {}Point(int a, int b) : x(a), y(b) {}};struct Node{int v, step;Node(int a, int b) : v(a), step(b) {}};int V;// 所有的G和Y的总数int swit;// 所有开关的，压缩到一个int里int n, m;char arr[22][22];Point points[22];//第i个节点的坐标int id[22][22];// 坐标为(i,j)的节点的编号vector<int> edges[333];void addEdge(int from, int to) {edges[from].push_back(to);}// 根据离散化后的坐标返回节点的编号int getID(int v) {return id[v / m][v % m];}int dist[22][22];void spfa(int s) {int sid = getID(s);for (int i = 0; i <= V; i++)dist[sid][i] = INF;dist[sid][sid] = 0;Node now = Node(s, 0);int done[333];memset(done, 0, sizeof(done));done[s] = 1;queue<Node> q;q.push(now);while (!q.empty()) {Node current = q.front();q.pop();int cid = getID(current.v);if (cid != -1) dist[sid][cid] = current.step;for (int i = 0; i < edges[current.v].size(); i++) {int v = edges[current.v][i];if (!done[v]) {done[v] = 1;q.push(Node(v, current.step + 1));}}}}void createGraph() {for (int i = 0; i < n; i++) {for (int j = 0; j < m; j++) {if (arr[i][j] == 'D') continue;if (arr[i][j] == 'F') {points[0] = Point(i, j);id[i][j] = 0;}if (arr[i][j] == 'G') {points[V] = Point(i, j);id[i][j] = V++;}if (arr[i][j] == 'Y') {points[V] = Point(i, j);id[i][j] = V;swit |= (1 << V++);}// 离散化坐标，每个点与四周能连的连边if (i > 0 && arr[i - 1][j] != 'D')addEdge(i * m + j, (i - 1) * m + j);if (i < n - 1 && arr[i + 1][j] != 'D')addEdge(i * m + j, (i + 1) * m + j);if (j > 0 && arr[i][j - 1] != 'D')addEdge(i * m + j, i * m + j - 1);if (j < m - 1 && arr[i][j + 1] != 'D')addEdge(i * m + j, i * m + j + 1);}}}int vis[22];int mid;  // 二分的初始电量bool dfs(int v, int energy, int stat) {if ((stat & swit) == swit) return true;// 遍历了所有的开关for (int i = 0; i < V; i++) {if (vis[i] || dist[v][i] == -1 || energy < dist[v][i]) continue;vis[i] = 1;if (arr[points[i].x][points[i].y] == 'G') {// 有充电器的话下一次的电量就是初始电量if (dfs(i, mid, stat | (1 << i))) return true;} else {// 否则消耗一定的电if (dfs(i, energy - dist[v][i], stat | (1 << i))) return true;}vis[i] = 0;}return false;}int getAns() {int l = 0, r = 222222;int ans;// 二分答案while (l <= r) {mid = l + r >> 1;memset(vis, 0, sizeof(vis));vis[0] = 1;if (dfs(0, mid, 1)) {ans = mid;r = mid - 1;} else l = mid + 1;}return ans;}int main() {while (~scanf("%d %d", &n, &m) && n && m) {memset(arr, 0, sizeof(arr));memset(points, 0, sizeof(points));memset(id, -1, sizeof(id));for (int i = 0; i <= n * m; i++) edges[i].clear();V = 1;swit = 0;for (int i = 0; i < n; i++)scanf("%s", arr[i]);createGraph();// 球两两之间最短路for (int i = 0; i < V; i++) {spfa(points[i].x * m + points[i].y);}bool flag = false;for (int i = 1; i < V; i++) {// 如果有Y无法到达就输出-1if (arr[points[i].x][points[i].y] == 'Y' && dist[0][i] == INF) {printf("-1\n");flag = true;break;}}if (flag) continue;int ans = getAns();cout << ans << endl;}return 0;}