hdu2612

Find a way Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 12932    Accepted Submission(s): 4141 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

#include<cstdio> #include<iostream> #include<cstring> #include<queue> using namespace std; int dir[4][2]={1,0, -1,0, 0,1, 0,-1}; int ori[211][211]; int map[211][211]; int x_s1,y_s1; int x_s2,y_s2; int n,m; int ans[211][211]; struct node { int x,y; int step; }; int min(int a,int b) { return a>b?b:a; } int judge(int x,int y) { if(x<0 || x>=n || y<0 || y>=m) return 1; if(map[x][y]==1) return 1; return 0; } void BFS(int x,int y) { queue<node>q; node cur,next; int i; cur.x=x; cur.y=y; cur.step=0; map[cur.x][cur.y]=1; q.push(cur); while(!q.empty()) { cur=q.front(); q.pop(); if(ori[cur.x][cur.y]==2) ans[cur.x][cur.y]+=cur.step; for(i=0;i<4;i++) { next.x=cur.x+dir[i][0]; next.y=cur.y+dir[i][1]; if(judge(next.x,next.y)) continue; next.step=cur.step+1; map[next.x][next.y]=1; q.push(next); } } } int main() { int i,l; char str[211]; int Ans; while(scanf("%d%d",&n,&m)!=-1) { memset(ori,0,sizeof(ori)); memset(ans,0,sizeof(ans)); for(i=0;i<n;i++) { scanf("%s",str); for(l=0;str[l];l++) { if(str[l]=='Y') {x_s1=i;y_s1=l;} else if(str[l]=='M'){x_s2=i;y_s2=l;} else if(str[l]=='.')ori[i][l]=0; else if(str[l]=='#')ori[i][l]=1; else if(str[l]=='@')ori[i][l]=2; } } for(i=0;i<n;i++) for(l=0;l<m;l++) map[i][l]=ori[i][l]; BFS(x_s1,y_s1); for(i=0;i<n;i++) for(l=0;l<m;l++) map[i][l]=ori[i][l]; BFS(x_s2,y_s2); Ans=11111111; for(i=0;i<n;i++) { for(l=0;l<m;l++) { if(ans[i][l]) { Ans=min(Ans,ans[i][l]); } } } printf("%d\n",Ans*11); } return 0; }