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.
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. XXXXX1 5 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) FINISH It 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) FINISH God please help our poor hero. FINISH
Author
Ignatius.L
思路：
看见图就知道是搜索。。。。。
一开始想用DFS因为DFS好写，但因为需要求最短路径，DFS不好处理，所以直接用BFS
唯一难一点的地方就是如何记录路径
因为当一个点第一次被BFS时的路径一定是到起点的最短路径，因此只需记录下此节点的前驱，最后打印时由终点回溯即可
当然，也可以从终点开始搜索，打印路径时方便一些
lrj的算法竞赛入门经典也有类似篇幅介绍BFS记录路径
代码：
//HDU 1026 //光搜 + 优先队列，需要记录路径 #include<stdio.h> #include<queue> #include<iostream> #include<string.h> using namespace std; const int MAXN = 110; struct node { int x, y; int time; friend bool operator<(node a, node b) //time小的优先级高 { return a.time > b.time; } }; priority_queue<node>que; //优先队列 struct cmap { int nx, ny; //记录前驱，用来记录路径 char c; } map[MAXN][MAXN]; //记录map int n, m; int fight[MAXN][MAXN], mark[MAXN][MAXN]; int bfs()//从目标点开始搜索，搜索到起点。记录搜索的前驱，就记录下路径了 { int k; int dir[4][2] = {{1, 0}, { -1, 0}, {0, 1}, {0, -1}}; node now, next; while (!que.empty()) que.pop();//初始化 now.x = n - 1; now.y = m - 1; if (map[now.x][now.y].c >= '1' && map[now.x][now.y].c <= '9') { now.time = map[n - 1][m - 1].c - '0'; fight[now.x][now.y] = map[now.x][now.y].c - '0'; } else now.time = 0; que.push(now); while (!que.empty()) { now = que.top(); que.pop(); if (now.x == 0 && now.y == 0) return now.time; for (k = 0; k < 4; k++) { next.x = now.x + dir[k][0]; next.y = now.y + dir[k][1]; if (next.x >= 0 && next.x < n && next.y >= 0 && next.y < m && !mark[next.x][next.y] && map[next.x][next.y].c != 'X') { if (map[next.x][next.y].c >= '1' && map[next.x][next.y].c <= '9') { next.time = now.time + map[next.x][next.y].c - '0' + 1; fight[next.x][next.y] = map[next.x][next.y].c - '0'; } else next.time = now.time + 1; que.push(next); map[next.x][next.y].nx = now.x; map[next.x][next.y].ny = now.y; mark[next.x][next.y] = 1; } } } return -1; } int main() { int i, j, flag, x, y, tx, ty, sec; while (scanf("%d%d", &n, &m) != EOF) { for (i = 0; i < n; i++) for (j = 0; j < m; j++) { scanf(" %c", &map[i][j].c); //前面一个空格很关键 mark[i][j] = fight[i][j] = 0; } mark[n - 1][m - 1] = 1; flag = bfs(); if (flag != -1) { printf("It takes %d seconds to reach the target position, let me show you the way. ", flag); sec = 1, x = y = 0; while (sec != flag + 1) { printf("%ds:(%d,%d)->(%d,%d) ", sec++, x, y, map[x][y].nx, map[x][y].ny); for (i = 0; i < fight[map[x][y].nx][map[x][y].ny]; i++) printf("%ds:FIGHT AT (%d,%d) ", sec++, map[x][y].nx, map[x][y].ny); tx = map[x][y].nx; ty = map[x][y].ny; x = tx; y = ty; } } else printf("God please help our poor hero. "); printf("FINISH "); } return 0; }