HDU 2612 Find a way（找条路）

HDU 2612 Find a way（找条路）

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

 

Problem Description - 题目描述

　　Pass a year learning in Hangzhou, yifenfei arrival hometown Ningbo at finally. Leave Ningbo one year, yifenfei have many people to meet. Especially a good friend Merceki.
　　Yifenfei’s home is at the countryside, but Merceki’s home is in the center of city. So yifenfei made arrangements with Merceki to meet at a KFC. There are many KFC in Ningbo, they want to choose one that let the total time to it be most smallest.
　　Now give you a Ningbo map, Both yifenfei and Merceki can move up, down ,left, right to the adjacent road by cost 11 minutes.

经过一年在杭州的学习，yifenfei终于回到了家乡宁波。离开宁波一年，yifenfei有许多人需要拜访。特别是好基友Merceki。
Yifenfei家住郊区，然而Merceki则住在市中心。因此yifenfei打算约Merceki在KFC面基，不过他们要选一家到达所花费的时间和最短的店。
现在给你一张宁波地图，yifenfei和Merceki可以花费11分钟分别进行上下左右移动。

Input - 输入

　　The input contains multiple test cases.
　　Each test case include, first two integers n, m. (2<=n,m<=200).
　　Next n lines, each line included m character.
　　　　‘Y’ express yifenfei initial position.
　　　　‘M’    express Merceki initial position.
　　　　‘#’ forbid road;
　　　　‘.’ Road.
　　　　‘@’ KCF

输入有多组测试用例。
每组用例开头为两个整数n，m。（2<=n,m<=200）
随后n行，每行有m个字符。
‘Y’表示yifenfei 的初始位置。
‘M’表示Merceki 的初始位置。
‘#’障碍物
‘.’道路
‘@’ KCF

Output - 输出

　　For each test case output the minimum total time that both yifenfei and Merceki to arrival one of KFC.You may sure there is always have a KFC that can let them meet.

对于每个测试用例，输出yifenfei 与Merceki 到达某个KFC最少的二者时间和。至少存在一个KFC让他们相遇。

Sample Input - 输入样例

4 4
Y.#@
....
.#..
@..M
4 4
Y.#@
....
.#..
@#.M
5 5
Y..@.
.#...
.#...
@..M.
#...#

Sample Output - 输出样例

66
88
66

 

题解
　　BFS套路，扫两遍出结果。
　　需要注意无法达到的KFC，可以通过处理最大值来规避。

 

代码 C++

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define INF 0x01010101
struct Point{
    int y, x;
}st1, st2, kfc[405], now, nxt;
std::queue<Point> q;
int data[205][205], iK, len[405], fx[8] = { 1, 0, -1, 0, 0, 1, 0, -1 };
char map[205][205];
void BFS(char c, char b){
    int i, tmp;
    while (!q.empty()){
        now = q.front(); q.pop(); tmp = data[now.y][now.x] + 1;
        for (i = 0; i < 8; i += 2){
            nxt.y = now.y + fx[i]; nxt.x = now.x + fx[i + 1];
            if (map[nxt.y][nxt.x] == c){
                data[nxt.y][nxt.x] = tmp; map[nxt.y][nxt.x] = b;
                q.push(nxt);
            }
        }
    }
    for (i = 0; i < iK; ++i) len[i] += data[kfc[i].y][kfc[i].x];
}
int main(){
    int n, m, i, j, opt;
    while (~scanf("%d%d ", &n, &m)){
        memset(map, '#', sizeof map); memset(len, 0, sizeof len);
        iK = 0; opt = INF;
        for (i = 1; i <= n; ++i) gets(&map[i][1]);
        for (i = 1; i <= n; ++i) for (j = 1; j <= m; ++j){
            switch (map[i][j]){
            case 'Y': st1 = { i, j }; map[i][j] = '.'; break;
            case 'M': st2 = { i, j }; map[i][j] = '.'; break;
            case '@': kfc[iK++] = { i, j }; map[i][j] = '.';
            }
        }

        memset(data, INF, sizeof data);
        q.push(st1); data[st1.y][st1.x] = 0; map[st1.y][st1.x] = '*';
        BFS('.', '*');

        memset(data, INF, sizeof data);
        q.push(st2); data[st2.y][st2.x] = 0; map[st2.y][st2.x] = 'X';
        BFS('*', 'X');

        for (i = 0; i < iK; ++i) opt = std::min(opt, len[i]);
        printf("%d
", opt * 11);
    }
    return 0;
}