HDU

Find a way

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 16184 Accepted Submission(s): 5194

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

Author

yifenfei

Source

奋斗的年代

Recommend

yifenfei

题意：Y和M在两个不同起点，他们要到KFC集合，路上有多家KFC，问到哪家KFC能使他们的步数和最少？

思路：两边bfs，分别存取Y和M到各家KFC的步数，相加求和，枚举各KFC输出最小值。

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;

char a[205][205];
int b[205][205],c[205][205];
int t[4][2]={{1,0},{0,1},{-1,0},{0,-1}};

struct Node{
    int x,y,s;
}node;

int main()
{
    int n,m,yx,yy,mx,my,i,j;
    while(~scanf("%d%d",&n,&m)){
        memset(c,0,sizeof(c));
        queue<Node> q;
        for(i=0;i<n;i++){
            getchar();
            scanf("%s",a[i]);
            for(j=0;j<m;j++){
                if(a[i][j]=='Y'){
                    yx=i;
                    yy=j;
                }
                if(a[i][j]=='M'){
                    mx=i;
                    my=j;
                }
            }
        }
        memset(b,0,sizeof(b));
        b[yx][yy]=1;
        node.x=yx;
        node.y=yy;
        node.s=0;
        q.push(node);
        while(q.size()){
            for(i=0;i<4;i++){
                int tx=q.front().x+t[i][0];
                int ty=q.front().y+t[i][1];
                if(tx<0||ty<0||tx>=n||ty>=m) continue;
                if(a[tx][ty]=='#'||b[tx][ty]==1) continue;
                b[tx][ty]=1;
                if(a[tx][ty]=='@'){
                    c[tx][ty]=q.front().s+11;
                }
                node.x=tx;
                node.y=ty;
                node.s=q.front().s+11;
                q.push(node);
            }
            q.pop();
        }
        memset(b,0,sizeof(b));
        b[mx][my]=1;
        node.x=mx;
        node.y=my;
        node.s=0;
        q.push(node);
        while(q.size()){
            for(i=0;i<4;i++){
                int tx=q.front().x+t[i][0];
                int ty=q.front().y+t[i][1];
                if(tx<0||ty<0||tx>=n||ty>=m) continue;
                if(a[tx][ty]=='#'||b[tx][ty]==1) continue;
                b[tx][ty]=1;
                if(a[tx][ty]=='@'){
                    c[tx][ty]+=q.front().s+11;
                }
                node.x=tx;
                node.y=ty;
                node.s=q.front().s+11;
                q.push(node);
            }
            q.pop();
        }
        int min=1000000000;
        for(i=0;i<n;i++){
            for(j=0;j<m;j++){
                if(c[i][j]<min&&c[i][j]!=0) min=c[i][j];
            }
        }
        printf("%d
",min);
    } 
    return 0;
}