HDU2612 find a way

题目链接：Find a way

bfs水题。

/*
  只有两个人啊。分别以两个人为起点bfs，计算出每个人到每个KFC 的时间。两个人都能到达的KFC的较大时间的最小值、就是ans。好水。T_T
 */

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <string>
#include <queue>
#define maxn 1000000
using namespace std;

char mp[210][210];
int vis[210][210];
int step[210][210], step1[210][210], step2[210][210];
int n, m;

struct Node{
    int x, y;
}now, temp, a, b;

queue<Node>que;

int dir[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};

bool check(Node a) {
    if (a.x >= 0 && a.x < n && a.y >= 0 && a.y < m && !vis[temp.x][temp.y] && mp[temp.x][temp.y] != '#')
        return true;
    return false;
}

void bfs1(Node t) {
    while(!que.empty()) {
        que.pop();
    }
    memset(vis, 0, sizeof(vis));
    que.push(t);
    vis[t.x][t.y] = 1;
    step1[t.x][t.y] = 0;

    while(!que.empty()) {
        now = que.front();
        que.pop();
        for (int i=0; i<4; ++i) {
            temp.x = now.x + dir[i][0];
            temp.y = now.y + dir[i][1];
            if (check(temp)) {
                vis[temp.x][temp.y] = 1;
                step1[temp.x][temp.y] = step1[now.x][now.y] + 1;
                que.push(temp);
            }
        }
    }
}

void bfs2(Node t) {
     while(!que.empty()) {
        que.pop();
    }
    memset(vis, 0, sizeof(vis));
    que.push(t);
    vis[t.x][t.y] = 1;
    step2[t.x][t.y] = 0;

    while(!que.empty()) {
        now = que.front();
        que.pop();
        for (int i=0; i<4; ++i) {
            temp.x = now.x + dir[i][0];
            temp.y = now.y + dir[i][1];
            if (check(temp)) {
                vis[temp.x][temp.y] = 1;
                step2[temp.x][temp.y] = step2[now.x][now.y] + 1;
                que.push(temp);
            }
        }
    }
}

int main() {
    while(cin >> n >> m) {
        memset(vis, 0, sizeof(vis));
        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                step[i][j] = maxn;
                step1[i][j] = maxn;
                step2[i][j] = maxn;
            }
        }
        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                cin >> mp[i][j];
                if (mp[i][j] == 'M') {
                    a.x = i, a.y = j;
                }
                else if (mp[i][j] == 'Y') {
                    b.x = i, b.y = j;
                }
            }
        }

        bfs1(a);
        bfs2(b);
        int ans = maxn;
        for (int i=0; i<n; ++i) {
            for (int j=0; j<m; ++j) {
                if (mp[i][j] == '@') {
                    step[i][j] = step1[i][j] + step2[i][j];
                }
                ans = min(ans, step[i][j]);
            }
        }
        cout << ans*11 << endl;
    }
    return 0;
}