hdu-1242 Rescue DFS解法

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25764    Accepted Submission(s): 9126

Problem Description

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

 

Input

First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend. 

Process to the end of the file.

 

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life." 

 

Sample Input

7 8#.#####.#.a#..r.#..#x.....#..#.##...##...#..............

 

Sample Output

13
题目解法使用dfs就好了，不过需要注意一个坑点，不要被它的示例数据给迷惑了，注意看题目中说的是Angel's friends！也就是说朋友可能不止一个，不止一个'r'，那要记录各个'r'所在的位置再去dfs，那就很麻烦了，所以这题可以反着来，用Angel的位置去dfs，找到各个朋友的位置，再取最少步数的一个就好了，具体代码如下：
////  main.cpp//  hdu1242////  Created by Morris on 16/7/21.//  Copyright © 2016年 Morris. All rights reserved.//#include <cstdio> #include <cstring> #include <climits>#define MAX_SZ 1000namespace {    using std::scanf;    using std::printf;    using std::memset;}int n, m;int min;//int flag;int dir[4][2] = { { 0, -1 }, { 0, 1 }, { -1, 0 }, { 1, 0 } };int visit[MAX_SZ][MAX_SZ];char map[MAX_SZ][MAX_SZ];void dfs(int x, int y, int cnt);int main(int argc, const char *argv[]){    int i, j;    while (~scanf("%d%d%*c", &n, &m)) {        int sx = 0, sy = 0;        for (i = 0; i < n; ++i) {            for (j = 0; j < m; ++j) {                scanf("%c", &map[i][j]);                if (map[i][j] == 'a') {                    sx = i;                    sy = j;                }            }                        getchar();        }                memset(visit, 0, sizeof(visit));        min = INT_MAX;        dfs(sx, sy, 0);                if (min != INT_MAX) {            printf("%d\n", min);        }        else {            printf("Poor ANGEL has to stay in the prison all his life.\n");        }    }    return 0;}void dfs(int x, int y, int cnt){    if (x < 0 || y < 0 || x >= n || y >= m) {        return ;    }    if (visit[x][y] == 1) {        return ;    }    if (cnt > min) {        return ;    }    if (map[x][y] == '#') {        return ;    }    if (map[x][y] == 'x') {        ++cnt;    }    if (map[x][y] == 'r') {        min = cnt < min ? cnt : min;        return ;    }        int i, tx, ty;    for (i = 0; i < 4; ++i) {        tx = x + dir[i][0];        ty = y + dir[i][1];                visit[x][y] = 1;        dfs(tx, ty, cnt + 1);        visit[x][y] = 0;    }}