[kuangbin带你飞]专题一 简单搜索-N - Find a way HDU - 2612

Find a way

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 20623    Accepted Submission(s): 6674

Problem Description

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.

 

Input

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

 

Output

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.

 

Sample Input

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

 

Sample Output

66 88 66

 

Author

yifenfei

 

Source

奋斗的年代

 

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题解：

        挺简单的题目，不知道为什么总是错，放在这里。以后再看。

     大神AC代码

#include <stdio.h>
#include <string.h>
#include <queue>
#include <algorithm>
#define inf 0x6ffffff
using namespace std;
char map [202][202];    // 地图
int vis  [202][202];    //  标记数组
int flag1[202][202];    //记录M到达任意KFC的时间
int flag [202][202];    //记录Y到达任意KFC的时间
int n,m;
int x1,y1,x2,y2;
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};

struct node{
	int x,y,step;
};

bool check(int x,int y){
	if(x < 0 ||y < 0 ||x >= n ||y >= m ||map[x][y] == '#' ||vis[x][y]) //检查是否越界，是否已经搜过
		return 0;
	return 1;
}

void bfs(int x,int y,int a[][202]) //进入坐标和记录数组
{
	int i;
	node st,ed;
	queue<node>q;
	st.x = x;
	st.y = y;
	st.step = 0;
	q.push(st);
	while(!q.empty()){
		st = q.front();
		q.pop();
		for(i = 0;i < 4;i ++){
			ed.x = st.x + dir[i][0];
			ed.y = st.y + dir[i][1];
			if(!check(ed.x,ed.y))
				continue;
			ed.step = st.step+1;
			vis[ed.x][ed.y] = 1;
		    if(map[ed.x][ed.y] == '@'){
				a[ed.x][ed.y] = ed.step;
			}
			q.push(ed);
		}
	}
}
int main()
{
	int i,j;

	while(~scanf("%d%d",&n,&m))
	{
		memset(flag,0,sizeof(flag));
		memset(vis,0,sizeof(vis));
		memset(flag1,0,sizeof(flag1));//全部初始化
		for(i = 0;i < n;i ++)
			scanf("%s",map[i]);
		for(i = 0;i < n;i ++)
			for(j = 0;j < m;j ++){
				if(map[i][j] == 'Y'){
					x1 = i;
					y1 = j;
				}
				if(map[i][j] == 'M'){
					x2 = i;
					y2 = j;
				}
			}
			vis[x1][y1] = 1;
			bfs(x1,y1,flag);//一遍BFS之后
			memset(vis,0,sizeof(vis));//初始化
			vis[x2][y2] = 1;
			bfs(x2,y2,flag1);  //再跑一遍BFS。
			int min = inf;
			for(i = 0;i < n;i ++)
				for(j = 0;j < m;j ++) //遍历整个地图、
                {
					if(min > flag[i][j] + flag1[i][j] && flag[i][j] && flag1[i][j])//有值的地方说明有KFC，找出两个人同时到达一个KFC的最短时间
						min = flag[i][j] + flag1[i][j];
				}
				printf("%d\n",min*11);//每走一步为11分钟。
	}
}

我的含有未知BUG的代码

#include <bits/stdc++.h>
#define mp(a,b) make_pair(a,b)
using namespace std;

typedef pair<int,int> Pi;
const int MAXN = 307;
const int INF = (int)0x3f3f3f3f;

int n,m;
Pi Mer,Yi;
vector<Pi> KFC;
int MeDis [MAXN][MAXN];
int Yidis [MAXN][MAXN];
char pic  [MAXN][MAXN];
int visM  [MAXN][MAXN];
int visY  [MAXN][MAXN];

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

void init(){
    memset(MeDis,0x3f,sizeof(MeDis));
    memset(Yidis,0x3f,sizeof(MeDis));
    memset(visM,0,sizeof(MeDis));
    memset(visY,0,sizeof(MeDis));
}

void BFS(){
    queue<Pi> qu;
    qu.push(Mer);
    visM[Mer.first][Mer.second] = 1;
    MeDis[Mer.first][Mer.second] = 0;

    while (!qu.empty()){
        Pi k = qu.front();
        qu.pop();

        for (int i = 0;i < 4;i ++){
            int nx = k.first + dir[i][0];
            int ny = k.second + dir[i][1];

            if (pic[nx][ny] == '#' || nx < 0 || nx >= n || ny < 0 || ny >= m || visM[nx][ny] )continue;
            visM[nx][ny] = 1;
            MeDis[nx][ny] = MeDis[k.first][k.second]+1;
            qu.push(mp(nx,ny));
        }
    }

    qu.push(Yi);
    visY[Yi.first][Yi.second] = 1;
    Yidis[Yi.first][Yi.second] = 0;

    while (!qu.empty()){
        Pi k = qu.front();
        qu.pop();

        for (int i = 0;i < 4;i ++){
            int nx = k.first + dir[i][0];
            int ny = k.second + dir[i][1];

            if (pic[nx][ny] == '#' || nx < 0 || nx >= n || ny < 0 || ny >= m || visY[nx][ny] )continue;
            visY[nx][ny] = 1;
            Yidis[nx][ny] = Yidis[k.first][k.second]+1;
            qu.push(mp(nx,ny));
        }
        //cout << qu.size() << endl;
    }
}

int main()
{
    ios::sync_with_stdio(false);

    while (cin >> n >> m){
        init();
        for (int i = 0;i < n;i ++){
            cin >> pic[i];
        }

        for (int i = 0;i < n;i ++){
            for (int j = 0;j < m;j ++){
                if (pic[i][j] == '@')  KFC.push_back(mp(i,j));
                else if (pic[i][j] == 'Y')  Yi.first = i,Yi.second = j;
                else if (pic[i][j] == 'M')  Mer.first = i,Mer.second = j;
            }
        }

        BFS();

        int MinNum = INF;
        for (int i = 0;i < KFC.size();i ++){
            int f = KFC[i].first;
            int s = KFC[i].second;
            if (Yidis[f][s] == INF || MeDis[f][s] == INF)continue;
            MinNum = min(MinNum,Yidis[f][s]+MeDis[f][s]);
        }

        cout << MinNum*11 << endl;

        for (int i = 0;i < n;i ++){
            for (int j = 0;j < m;j ++){
                cout << MeDis[i][j] << ',';
            }
            cout << endl;
        }

    }
}