UVA - 12569 Planning mobile robot on Tree (EASY Version) BFS

题意：有一颗n个节点的树，其中一个节点有机器人，指定m个节点有障碍物，指定终点，问最少需要移动多少步才能到达，移动过程中遇到障碍物需要将其移动到空地。

分析：bfs搜索，是每次都对障碍物进行移动，将障碍物移动到空闲地方，如果是机器人移动就需要更新位置。

障碍物位置用二进制储存，方便查找，更新。

# include<iostream>
# include<cstdio>
# include<cmath>
# include<map>
# include<queue>
# include<string>
# include<string.h>
#include<set>
#include<list>
# include<algorithm>
using namespace std;
const int maxn = 1 << 20;
struct node {int state, pos, pre, len;node(){}node(int _state,int _pos,int _pre,int _len):state(_state),pos(_pos),pre(_pre),len(_len){}
};
vector<int>G[20];
node q[maxn];
int vis[20][maxn];
int n, obnum, s, t;
int start,step;
void bfs(int ob) {int front = 0, rear = 1;node temp(ob, s, -1, 0);q[front] = temp;vis[s][ob] = 1;while (front < rear) {node u = q[front];if (u.pos == t) {//移动到终点就退出start = front;step = u.len;break;}for (int i = 0; i < n; i++) {if ((1 << i)&u.state) {//找到障碍物位置for (int j = 0; j < G[i].size(); j++) {int next = G[i][j];if ((1 << next)&u.state)continue;//找到空闲位置node ttd = u;ttd.pos = u.pos; ttd.state = ((u.state|(1<<next))^(1<<i));if (ttd.pos == i)ttd.pos = next;//如果是机器人移动，就更新位置if (!vis[ttd.pos][ttd.state]) {vis[ttd.pos][ttd.state] = 1;ttd.len++;ttd.pre = front;q[rear++] = ttd;}}}}front++;}
}
void result(int u) {if (q[u].pre != 0)result(q[u].pre);//从前往后输出int last = q[q[u].pre].state;int cur = q[u].state;;int a, b;a = last ^ (last&cur);//因为每次只移动一步b = cur ^ (last&cur);int aa, bb;for (int i = 0; i < n; i++) {if ((1 << i)&a)aa = i;if ((1 << i)&b)bb = i;}cout << aa + 1 << " " << bb + 1 << endl;
}
int main() {int kase;cin >> kase;for (int k = 0; k < kase; k++) {for (int i = 0; i < 20; i++)G[i].clear();memset(vis, 0, sizeof(vis));step = -1;cin >> n >> obnum >> s >> t;int obstacle = 0;s--, t--;for (int i = 0; i < obnum; i++) { int l;cin >> l;l--;obstacle |= (1<<l);}for (int i = 0; i < n - 1; i++) { int temp,temp1; cin >> temp>>temp1;G[temp-1].push_back(temp1-1);G[temp1-1].push_back(temp-1);}obstacle |= (1 << s);bfs(obstacle);cout << "Case " << k + 1 << ": ";if (step == -1) {cout << "-1" << endl;}else {cout << step << endl;result(start);}}return 0;
}

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UVA - 12569 Planning mobile robot on Tree (EASY Version) BFS

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