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最新推荐文章于 2020-07-17 23:25:50 发布

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最新推荐文章于 2020-07-17 23:25:50 发布

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本文链接：https://blog.youkuaiyun.com/acdalao/article/details/96650814

博客围绕HDU2612题目展开，描述地图上小Y和小M从家出发去KFC碰面，每单位移动需11分钟。输入地图信息，输出碰面所需最短时间。题解是分别以小Y和小M为起点进行两次BFS，预处理到所有KFC时间，再遍历找最小和。

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传送门HDU2612

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.

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.

样例

Input

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

Output
66
88
66

题解

题意：地图上有多个KFC，小Y和小M一起从家出发前往一家KFC碰面，每单位可以往上下左右走，求在哪个KFC碰面所需时间最短，输出最短时间
分别以小Y和小M为起点跑两次BFS，预处理处理他们到所有KFC所需时间，然后遍历找最小和即可。

Code

#include<bits/stdc++.h>
#define INIT(a,b) memset(a,b,sizeof(a))
#define LL long long
using namespace std;
const int inf=0x3f3f3f3f;
const int N=2e2+7;
const int mod=1e9+7;
char mp[N][N];
int vis[N][N];
int ans[N][N][2];
int n,m,sx,sy,ex,ey;
int rec[4][2]={1,0, -1,0, 0,1, 0,-1};
struct node{
    int i,j,time;
};
bool judge(int x,int y){
    if(x<0||x>=n||y<0||y>=m||vis[x][y]||mp[x][y]=='#')return false;
    return true;
}
void bfs(int x,int y,int t){
    INIT(vis,0);
    queue<node> que;
    que.push((node){x,y,0});
    vis[x][y]=1;
    while(!que.empty()){
        node q=que.front();que.pop();
        if(mp[q.i][q.j]=='@') ans[q.i][q.j][t]=q.time * 11;
        for(int k=0;k<4;k++){
            int tx=q.i+rec[k][0],ty=q.j+rec[k][1];
            if(judge(tx,ty)){
                vis[tx][ty]=1;
                que.push((node){tx,ty,q.time+1});
            }
        }
    }
}
int main(){
    while(cin>>n>>m){
        INIT(ans,inf);
        getchar();
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cin>>mp[i][j];
                if(mp[i][j]=='Y') sx=i,sy=j;
                if(mp[i][j]=='M') ex=i,ey=j;
            }
        }
        bfs(sx,sy,0);
        bfs(ex,ey,1);
        int res=inf;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                res=min(ans[i][j][0]+ans[i][j][1],res);
            }
        }
        cout<<res<<endl;
    }
	return 0;
}