HDU-P2612:Find A Way[广度优先搜索]

题目

思路

题目的大致内容很容易理解，进行两次广度优先搜索找到两次路径和最小的KFC的位置即可。

代码

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>
#include<math.h>
#include<set>
using namespace std;
#define INF 1e9
typedef long long ll;
char ch[300][300];
int vis[300][300],n,m;
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
struct point
{
    int x,y,step;
}start1,start2;
int node[205][205],dis[3][40005];
///node用来标记KFC的位置，dis用来记录Y到所有KFC、M到所有KFC的距离
void bfs(int t)
{
    point start;
    if(t == 1) start = start1;///1代表Y
    else start = start2;///2代表M
    memset(vis,0,sizeof(vis));
    start.step = 0;
    vis[start.x][start.y] = 1;
    point cur,nxt;
    queue<point> q;
    q.push(start);
    while(!q.empty())
    {
        cur = q.front();
        q.pop();
        for(int i = 0; i < 4; i++)
        {
            nxt.x = cur.x+dir[i][0];
            nxt.y = cur.y+dir[i][1];
            nxt.step = cur.step + 1;
            if(nxt.x>=0 && nxt.x<n && nxt.y>=0 && nxt.y<m && vis[nxt.x][nxt.y] == 0 && ch[nxt.x][nxt.y] != '#')
            {
                vis[nxt.x][nxt.y] = 1;
                ///碰到一个KFC，记录下到它的距离
                if(ch[nxt.x][nxt.y] == '@') dis[t][node[nxt.x][nxt.y]] = nxt.step;
                q.push(nxt);
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        int cunt = 1;
        for(int i = 0; i < n; i++)
        {
            getchar();
            scanf("%s",ch[i]);
            for(int j = 0; j < m; j++)
            {
                if(ch[i][j] == 'Y')
                {
                    start1.x = i;
                    start1.y = j;
                }
                if(ch[i][j] == 'M')
                {
                    start2.x = i;
                    start2.y = j;
                }
                if(ch[i][j] == '@')
                {
                    node[i][j] = cunt;
                    cunt++;
                }
            }
        }
        bfs(1);///Y到各个KFC的距离
        bfs(2);///M到各个KFC的距离
        int minn = INF;
        ///找出一个两人花费总时间最少的KFC
        for(int i = 1; i < cunt; i++)
        {
            minn =  min(minn,dis[1][i]+dis[2][i]);
        }
        printf("%d\n",minn*11);
    }
    return 0;
}