hdu 4198:Quick out of the Harbour解题报告

hdu 4198:Quick out of the Harbour解题报告

Quick out of the Harbour Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1441    Accepted Submission(s): 575 Problem DescriptionCaptain Clearbeard decided to go to the harbour for a few days so his crew could inspect and repair the ship. Now, a few days later, the pirates are getting landsick(Pirates get landsick when they don't get enough of the ships' rocking motion. That's why pirates often try to simulate that motion by drinking rum.). Before all of the pirates become too sick to row the boat out of the harbour, captain Clearbeard decided to leave the harbour as quickly as possible.Unfortunately the harbour isn't just a straight path to open sea. To protect the city from evil pirates, the entrance of the harbour is a kind of maze with drawbridges in it. Every bridge takes some time to open, so it could be faster to take a detour. Your task is to help captain Clearbeard and the fastest way out to open sea.The pirates will row as fast as one minute per grid cell on the map. The ship can move only horizontally or vertically on the map. Making a 90 degree turn does not take any extra time.  InputThe first line of the input contains a single number: the number of test cases to follow. Each test case has the following format:1. One line with three integers, h, w (3 <= h;w <= 500), and d (0 <= d <= 50), the height and width of the map and the delay for opening a bridge.2.h lines with w characters: the description of the map. The map is described using the following characters:—"S", the starting position of the ship.—".", water.—"#", land.—"@", a drawbridge.Each harbour is completely surrounded with land, with exception of the single entrance.  OutputFor every test case in the input, the output should contain one integer on a single line: the travelling time of the fastest route to open sea. There is always a route to open sea. Note that the open sea is not shown on the map, so you need to move outside of the map to reach open sea.  Sample Input26 5 7######S..##@#.##...##@####.###4 5 3######S#.##@..####@#  Sample Output1611   这个题大概的意思就是一艘船要从S点出发，到达唯一出口的最短时间，只能走'.'（有水的地方）或者'@'标记的地方,不能走'#'标记的地方,因为有'.'花费的时间为1,'@'花费的时间为第三个参数c+1，所以考虑优先队列。 下面是暑假培训c++代码： #include #include #include #include using namespace std; char mp[502][502]; int vis[502][502]; int dir[][2]= {{1,0},{-1,0},{0,1},{0,-1}}; int n,m,c,ex,ey; struct node { int x,y,step; }; bool operator < (node t1,node t2) { return t1.step>t2.step; } priority_queueq; int bfs() { node t,next; int tx; while(!q.empty()) { t=q.top(); q.pop(); if(t.x==ex&&t.y==ey) return t.step; for(int i=0; i<4; i++) { next.step=t.step; next.x=t.x+dir[i][0]; next.y=t.y+dir[i][1]; if(next.x<0||next.x>=n||next.y<0||next.y>=m||vis[next.x][next.y]||mp[next.x][next.y]=='#') continue; if(mp[next.x][next.y]=='@') next.step+=(c+1); if(mp[next.x][next.y]=='.') next.step++; vis[next.x][next.y]=1; q.push(next); } } return 0; } int main() { int tcase; scanf("%d",&tcase); while(tcase--) { node t; memset(mp,0,sizeof(mp)); memset(vis,0,sizeof(vis)); while(!q.empty()) q.pop(); scanf("%d%d%d",&n,&m,&c); for(int i=0; i{ int x, y, step; public node(){ } public node(int x,int y,int step){ this.x=x; this.y=y; this.step=step; } @Override public int compareTo(node o) { if(this.step>o.step) return 1; return -1; } } static int n, m, value; static int startx, starty, endx, endy; static char[][] map; static boolean[][] visit; static int dir[][] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } }; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int tcase = sc.nextInt(); while (tcase-- > 0) { n = sc.nextInt(); m = sc.nextInt(); value = sc.nextInt(); map = new char[n][m]; visit = new boolean[n][m]; for (int i = 0; i < n; i++) { String str = sc.next(); map[i] = str.toCharArray(); for (int j = 0; j < m; j++) { if (map[i][j] == 'S') { startx = i; starty = j; } if (map[i][j] != '#' && (i == 0 || i == n - 1 || j == 0 || j == m - 1)) { endx = i; endy = j; } } } System.out.println(BFS(startx,starty)); } } private static int BFS(int startx2, int starty2) { PriorityQueue q =new PriorityQueue(); node t = new node(startx2,starty2,1); //step初始值为1,S点也要算进去 q.add(t); visit[t.x][t.y]=true; while(!q.isEmpty()){ t = q.remove(); if(t.x==endx&&t.y==endy) { return t.step; } for(int i=0;i<4;i++){ int step = t.step; int x = t.x+dir[i][0]; int y = t.y+dir[i][1]; if(x<0||x>n-1||y<0||y>m-1||visit[x][y]||map[x][y]=='#') continue; if(map[x][y]=='@'){ step+=(value+1); } if(map[x][y]=='.'){ step++; } visit[x][y]=true; q.add(new node(x,y,step)); } } return 0; } }   问题代码： //问题代码 import java.util.PriorityQueue; import java.util.Scanner; public class Main { static class node implements Comparable{ int x, y, step; @Override public int compareTo(node o) { if(this.step>o.step) return 1; return -1; } } static int n, m, value; static int startx, starty, endx, endy; static int[][] map; static boolean[][] visit; static int dir[][] = { { 1, 0 }, { -1, 0 }, { 0, 1 }, { 0, -1 } }; public static void main(String[] args) { Scanner sc = new Scanner(System.in); int tcase = sc.nextInt(); while (tcase-- > 0) { n = sc.nextInt(); m = sc.nextInt(); value = sc.nextInt(); map = new int[n][m]; visit = new boolean[n][m]; for (int i = 0; i < n; i++) { String str = sc.next(); char[] x = str.toCharArray(); for (int j = 0; j < m; j++) { if (x[j] == 'S') { startx = i; starty = j; map[i][j] = 1; } if (x[j] == '@') { map[i][j] = value+ 1; } if (x[j] == '#') { map[i][j] = -1; } if (x[j] == '.') { map[i][j] = 1; } if (x[j] != '#' && (i == 0 || i == n - 1 || j == 0 || j == m - 1)) { endx = i; endy = j; map[i][j] = 0; } } } System.out.println(BFS(startx,starty)); } } private static int BFS(int startx2, int starty2) { PriorityQueue q =new PriorityQueue(); node t = new node(); t.x=startx; t.y=starty; t.step = 0; q.add(t); visit[t.x][t.y]=true; while(!q.isEmpty()){ t = q.remove(); //System.out.println(t.x+" "+t.y+" "+t.step); if(t.x==endx&&t.y==endy) { return t.step; } node t1 = new node(); for(int i=0;i<4;i++){ t1.step = t.step; t1.x = t.x+dir[i][0]; t1.y = t.y+dir[i][1]; if(t1.x<0||t1.x>n-1||t1.y<0||t1.y>m-1||visit[t1.x][t1.y]||map[t1.x][t1.y]==-1) continue; t1.step +=map[t1.x][t1.y]; //System.out.println("t1:"+t1.x+" "+t1.y+" "+t1.step); visit[t1.x][t1.y]=true; q.add(t1); //System.out.println(q.peek().x+" "+q.peek().y+" "+q.peek().step); } } return 0; } }

版权声明：本文内容由网络用户投稿，版权归原作者所有，本站不拥有其著作权，亦不承担相应法律责任。如果您发现本站中有涉嫌抄袭或描述失实的内容，请联系我们jiasou666@gmail.com 处理，核实后本网站将在24小时内删除侵权内容。