Stealing Harry Potter’s Precious
Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1982Accepted Submission(s): 931
Problem Description
Harry Potter has some precious. For example, his invisible robe, his wand and his owl. When Hogwarts school is in holiday, Harry Potter has to go back to uncle Vernon’s home. But he can’t bring his precious with him. As you know, uncle Vernon never allows such magic things in his house. So Harry has to deposit his precious in the Gringotts Wizarding Bank which is owned by some goblins. The bank can be considered as a N × M grid consisting of N × M rooms. Each room has a coordinate. The coordinates of the upper-left room is (1,1) , the down-right room is (N,M) and the room below the upper-left room is (2,1)….. A 3×4 bank grid is shown below:
Some rooms are indestructible and some rooms are vulnerable. Goblins always care more about their own safety than their customers’ properties, so they live in the indestructible rooms and put customers’ properties in vulnerable rooms. Harry Potter’s precious are also put in some vulnerable rooms. Dudely wants to steal Harry’s things this holiday. He gets the most advanced drilling machine from his father, uncle Vernon, and drills into the bank. But he can only pass though the vulnerable rooms. He can’t access the indestructible rooms. He starts from a certain vulnerable room, and then moves in four directions: north, east, south and west. Dudely knows where Harry’s precious are. He wants to collect all Harry’s precious by as less steps as possible. Moving from one room to another adjacent room is called a ‘step’. Dudely doesn’t want to get out of the bank before he collects all Harry’s things. Dudely is stupid.He pay you $1,000,000 to figure out at least how many steps he must take to get all Harry’s precious.
Input
There are several test cases. In each test cases: The first line are two integers N and M, meaning that the bank is a N × M grid(0<N,M <= 100). Then a N×M matrix follows. Each element is a letter standing for a room. ‘#’ means a indestructible room, ‘.’ means a vulnerable room, and the only ‘@’ means the vulnerable room from which Dudely starts to move. The next line is an integer K ( 0 < K <= 4), indicating there are K Harry Potter’s precious in the bank. In next K lines, each line describes the position of a Harry Potter’s precious by two integers X and Y, meaning that there is a precious in room (X,Y). The input ends with N = 0 and M = 0
Output
For each test case, print the minimum number of steps Dudely must take. If Dudely can’t get all Harry’s things, print -1.
Sample Input
2 3##@#.#12 24 4#@##….####….22 12 40 0
Sample Output
-15
Source
题目链接：?pid=4771
题目大意：一个人要在一个n*m的地图中找k个宝藏，求其收集完所有宝藏走过的最短路
题目分析：因为宝藏数不大于4，我们可以求出起点和所有宝藏两两之间的最短路，用BFS搜所出来，然后再枚举所有情况，找最短的那条即可，，排列数stl中有个叫next_permutation（）的函数相当好用啊
比如样例二起点是（1,2）宝藏在（2,1）和（2,3）那我们把起点设为A点，两个宝藏设为B,C，就两种可能ABC和ACB，再通过求出来的两点间的最短路得到这两种情况的总路程，最后取小的那个即为答案，代码狂丑….#include <cstdio>#include <cstring>#include <queue>#include <string>#include <algorithm>#include <iostream>using namespace std;int const MAX = 100 + 5;int map[MAX][MAX];int n, m, k, cnt;int dx[] = {1, 0, -1, 0};int dy[] = {0, 1, 0, -1};char s[105];struct Point //存点的信息{int x, y;int step;int id;}p[105];int ex, ey;int dis[105][105];bool vis[105][105];void BFS(Point pt, Point ept){vis[pt.x][pt.y] = true;Point now, t;queue <Point> q;q.push(pt);while(!q.empty()){now = q.front();q.pop();if(now.x == ex && now.y == ey) //记录两点间的最短路{dis[pt.id][ept.id] = now.step;dis[ept.id][pt.id] = now.step;return;}for(int i = 0; i < 4; i ++){t = now;t.x += dx[i];t.y += dy[i];t.step ++;if(t.x >= n || t.x < 0 || t.y >= m || t.y < 0 || vis[t.x][t.y] || map[t.x][t.y] == 0)continue;vis[t.x][t.y] = true;q.push(t);}}}int main(){while(scanf("%d %d", &n, &m) && (n + m)){bool f = false;int ans = 0;string next;char tmp[10];cnt = 0;memset(map, 0, sizeof(map));memset(dis, -1, sizeof(dis));for(int i = 0; i < n; i++){scanf("%s", s);for(int j = 0; j < m; j++){if(s[j] == '.')map[i][j] = 1;if(s[j] == '@'){map[i][j] = 3;p[cnt].x = i;p[cnt].y = j;p[cnt].id = cnt;p[cnt++].step = 0;}}}int bx, by;scanf("%d", &k);for(int i = 0; i < k; i++){scanf("%d %d", &bx, &by);if(map[bx – 1][by – 1] == 0)f = true;elsemap[bx – 1][by – 1] = 2;}for(int i = 0; i < n; i++)for(int j = 0; j < m; j++)if(map[i][j] == 2){p[cnt].x = i;p[cnt].y = j;p[cnt].id = cnt;p[cnt++].step = 0;} //将起点设置为0点方便后面的操作for(int i = 0; i < cnt; i++) //枚举任两点并搜其最短路径{for(int j = i + 1; j < cnt; j++){ex = p[j].x;ey = p[j].y;memset(vis, false, sizeof(vis));BFS(p[i],p[j]);}}for(int i = 0; i < cnt; i++) //如果存在宝藏不可达则说明无解{for(int j = 0; j < cnt; j++){if(i == j)continue;if(dis[i][j] == -1){f = true;break;}}if(f)break;}if(f){printf("-1\n");continue;}//计算每种情况下的总路程for(int i = 0; i < cnt – 1; i++)tmp[i] = i + 49;next.insert(0,tmp,0,cnt-1);int cur = 0;for(int i = 0; i < cnt – 1; i++){ans += dis[cur][next[i] – 48];cur = next[i] – 48;}while(next_permutation(next.begin(), next.end())) //举出所有排列{int re = 0;cur = 0;for(int i = 0; i < cnt – 1; i++){re += dis[cur][next[i] – 48];cur = next[i] – 48;}ans = min(ans, re);}printf("%d\n", ans);}}
旅游时最好的习惯：找个舒适的小店，挑张雅致的明信片，
