HDU

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=2612

题意
有两个人 要去一个城市中的KFC 一个城市中有多个KFC 求两个人到哪一个KFC的总时间最少 求出这个最少的总时间

思路
用BFS打一遍表 求出两个人每人到 每个 KFC 的时间 然后 遍历一下 更新答案

AC代码

#include <cstdio>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <cmath>
#include <climits>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <deque>
#include <vector>
#include <queue>
#include <string>
#include <map>
#include <stack>
#include <set>
#include <numeric>
#include <sstream>
#include <iomanip>
#include <limits>

#define CLR(a) memset(a, 0, sizeof(a))
#define pb push_back

using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair<string, int> psi;
typedef pair<string, string> pss;

const double PI = acos(-1);
const double E = exp(1);
const double eps = 1e-30;

const int INF = 0x3f3f3f3f;
const int maxn = 5e4 + 5;
const int MOD = 1e9 + 7;

int Move[][2]
{
    -1, 0, // up
     1, 0, // down
     0,-1, // left
     0, 1, // right
};

struct Node
{
    int x, y, step;
}tmp;

string G[205];

int v[205][205];
int dis[205][205][2];

int n, m;

int sx[2], sy[2];

queue <Node> q;

bool ok(int x, int y)
{
    if (x < 0 || x >= n || y < 0 || y >= m || G[x][y] == '#' || v[x][y] == 1)
        return false;
    return true;
}

void bfs(int vis)
{
    while (!q.empty())
    {
        int x = q.front().x;
        int y = q.front().y;
        int step = q.front().step;
        q.pop();
        if (G[x][y] == '@')
            dis[x][y][vis] = min(dis[x][y][vis], step);
        for (int i = 0; i < 4; i++)
        {
            tmp.x = x + Move[i][0];
            tmp.y = y + Move[i][1];
            tmp.step = step + 1;
            if (ok(tmp.x, tmp.y))
            {
                q.push(tmp);
                v[tmp.x][tmp.y] = 1;
            }
        }
    }
}

int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        for (int i = 0; i < n; i++)
        {
            cin >> G[i];
            for (int j = 0; j < m; j++)
            {
                if (G[i][j] == 'Y')
                {
                    sx[0] = i;
                    sy[0] = j;
                }
                else if (G[i][j] == 'M')
                {
                    sx[1] = i;
                    sy[1] = j;
                }
            }
        }
        memset(dis, 0x3f, sizeof(dis));
        for (int i = 0; i < 2; i++)
        {
            CLR(v);
            tmp.x = sx[i];
            tmp.y = sy[i];
            v[tmp.x][tmp.y] = 1;
            tmp.step = 0;
            while (!q.empty())
                q.pop();
            q.push(tmp);
            bfs(i);
        }
        int ans = INF;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (G[i][j] == '@')
                    ans = min(ans, dis[i][j][0] + dis[i][j][1]);
            }
        }
        cout << ans * 11 << endl;
    }
}