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hdu 1026 Ignatius and the Princess I (BFS+优先队列)
Ignatius and the Princess I
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11700 Accepted Submission(s): 3653
Special Judge
Problem Description
The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166‘s castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-top corner is (0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0), and the door to feng5166‘s room is at (N-1,M-1), that is our target. There are some monsters in the castle, if Ignatius meet them, he has to kill them. Here is some rules:
1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.
Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.
1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.
Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.
Input
The input contains several test cases. Each test case starts with a line contains two numbers N and M(2<=N<=100,2<=M<=100) which indicate the size of the labyrinth. Then a N*M two-dimensional array follows, which describe the whole labyrinth. The input is terminated by the end of file. More details in the Sample Input.
Output
For each test case, you should output "God please help our poor hero." if Ignatius can‘t reach the target position, or you should output "It takes n seconds to reach the target position, let me show you the way."(n is the minimum seconds), and tell our hero the whole path. Output a line contains "FINISH" after each test case. If there are more than one path, any one is OK in this problem. More details in the Sample Output.
Sample Input
5 6.XX.1...X.2.2...X....XX.XXXXX.5 6.XX.1...X.2.2...X....XX.XXXXX15 6.XX.....XX1.2...X....XX.XXXXX.
Sample Output
It takes 13 seconds to reach the target position, let me show you the way.1s:(0,0)->(1,0)2s:(1,0)->(1,1)3s:(1,1)->(2,1)4s:(2,1)->(2,2)5s:(2,2)->(2,3)6s:(2,3)->(1,3)7s:(1,3)->(1,4)8s:FIGHT AT (1,4)9s:FIGHT AT (1,4)10s:(1,4)->(1,5)11s:(1,5)->(2,5)12s:(2,5)->(3,5)13s:(3,5)->(4,5)FINISHIt takes 14 seconds to reach the target position, let me show you the way.1s:(0,0)->(1,0)2s:(1,0)->(1,1)3s:(1,1)->(2,1)4s:(2,1)->(2,2)5s:(2,2)->(2,3)6s:(2,3)->(1,3)7s:(1,3)->(1,4)8s:FIGHT AT (1,4)9s:FIGHT AT (1,4)10s:(1,4)->(1,5)11s:(1,5)->(2,5)12s:(2,5)->(3,5)13s:(3,5)->(4,5)14s:FIGHT AT (4,5)FINISHGod please help our poor hero.FINISH
题意:就是让你找到从(0,0)到(n-1,m-1)的最短路径,并把这条路径给输出来,若没有则输出God please help our poor hero,而且输出路径时如果遇到怪物,与它战斗了几秒也要输出来。
解题思路:用BFS+优先队列的方法,我们不断寻找下一个位置,并把耗时最少的又作为起点,寻找下一个位置,如果能找到,则第一次找到的一定是最短耗时的,但这一题有一个难点,就是难在如何存储我找到的最短路径。我们开一个结构体数组,当我们找到下一个位置的时候,在下一个节点中存储上一个节点的位置,这样我们就可以从终点递推来找到整条路,然后把它存储在另一个数组中,这样我们就直接输出这个数组就可以了。
贴出代码:
#include <stdio.h>#include <iostream>#include <queue>using namespace std;int n, m, mark, minx;int visited[105][105];int dir[4][2] = {0, -1, 1, 0, 0, 1, -1, 0};int Footx[10005], Footy[10005]; //作为转换的路径数组char map[105][105];struct foot //存储路径的结构体{ int x, y;}Foot[105][105];struct node //优先队列结构体,耗时少的优先{ int x, y; int time; friend bool operator < (node a, node b) { return a.time > b.time; }};int Judge(int x, int y) //判断这个点是否符合{ if(x<0 || x>=n || y<0 || y>=m) return 1; if(map[x][y] == ‘X‘ || visited[x][y]) return 1; return 0;}void BFS(int startx, int starty, int time) //BFS搜索最短路{ priority_queue <node> Q; struct node Node, temp; Node.x = startx; Node.y = starty; Node.time = 0; visited[startx][starty] = 1; Q.push(Node); while(!Q.empty()) { Node = Q.top(); Q.pop(); if(Node.x == n-1 && Node.y == m-1) { minx = Node.time; mark = 1; return; } for(int i = 0; i<4; i++) { temp.x = Node.x+dir[i][0]; temp.y = Node.y+dir[i][1]; temp.time = Node.time+1; if(Judge(temp.x, temp.y)) continue; if(map[temp.x][temp.y] == ‘.‘) { Foot[temp.x][temp.y].x = Node.x; //将上一个节点存储下来 Foot[temp.x][temp.y].y = Node.y; visited[temp.x][temp.y] = 1; Q.push(temp); } if(map[temp.x][temp.y]>=‘1‘ && map[temp.x][temp.y]<=‘9‘) { Foot[temp.x][temp.y].x = Node.x; Foot[temp.x][temp.y].y = Node.y; temp.time = temp.time+map[temp.x][temp.y]-‘1‘+1; visited[temp.x][temp.y] = 1; Q.push(temp); } } }}void Show(){ printf("It takes %d seconds to reach the target position, let me show you the way.\n", minx); int k = 0, i =minx, num; Footx[k] = n-1; Footy[k] = m-1; while(i--) //将存储下来的路径转换出来,我们就可以知道走过了那些点 { if(Footx[k] == 0 && Footy[k] == 0) break; k++; Footx[k] = Foot[ Footx[k-1] ][ Footy[k-1] ].x; Footy[k] = Foot[ Footx[k-1] ][ Footy[k-1] ].y; } num = 1; for(int j = k-1; j>=0; j--) //根据走过的这些点来进行输出 { if(map[ Footx[j] ][ Footy[j] ] == ‘.‘) printf("%ds:(%d,%d)->(%d,%d)\n", num++, Footx[j+1], Footy[j+1], Footx[j], Footy[j]); if(map[ Footx[j] ][ Footy[j] ]>=‘1‘ && map[ Footx[j] ][ Footy[j] ]<=‘9‘) { printf("%ds:(%d,%d)->(%d,%d)\n", num++, Footx[j+1], Footy[j+1], Footx[j], Footy[j]); for(int t = 1; t<=map[ Footx[j] ][ Footy[j] ]-‘0‘; t++) { printf("%ds:FIGHT AT (%d,%d)\n", num++, Footx[j], Footy[j]); } } }}int main(){ while(scanf("%d%d", &n, &m)!=EOF) { for(int i = 0; i<n; i++) { for(int j = 0; j<m; j++) { cin>>map[i][j]; visited[i][j] = 0; } } mark = 0; BFS(0, 0, 0); if(mark) Show(); else printf("God please help our poor hero.\n"); printf("FINISH\n"); } return 0;}
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