Find a way (BFS)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2612

 

BFS搜索  目标地  并记录下来  之后再判断两段路程之和   

 

代码:

  1 #include <stdio.h>
  2 #include <string.h>
  3 #include <math.h>
  4 #include <algorithm>
  5 #include <iostream>
  6 #include <ctype.h>
  7 #include <iomanip>
  8 #include <queue>
  9 #include <stdlib.h>
 10 using namespace std;
 11 
 12 #define INF 0x3f3f3f3f  
 13 #define N 220
 14 #define M 220
 15 int n,m;
 16 char mp[N][M];
 17 int vis1[N][M],vis2[N][M];
 18 int ans1[N][M],ans2[N][M];
 19 int dx[4]={0,0,1,-1};
 20 int dy[4]={1,-1,0,0};
 21 
 22 struct node
 23 {
 24     int x,y;
 25     int t;
 26 };
 27 
 28 node g1,g2;
 29 void getMp()
 30 {
 31     for(int i=0; i<n; i++){
 32         for(int j=0; j<m; j++){
 33             cin>>mp[i][j];
 34             if(mp[i][j]=='Y'){
 35                 g1.x=i;
 36                 g1.y=j;
 37                 g1.t=0;
 38                 vis1[i][j]=1;
 39             }
 40             if(mp[i][j]=='M'){
 41                 g2.x=i;
 42                 g2.y=j;
 43                 g2.t=0;
 44                 vis2[i][j]=1;
 45             }
 46         }
 47     }
 48 }
 49 
 50 void bfs1()
 51 {
 52     queue<node> q;
 53     while(!q.empty())
 54         q.pop();
 55     q.push(g1);
 56     while(!q.empty()){
 57         node f1;
 58         f1=q.front();
 59         q.pop();
 60         for(int i=0; i<4; i++){
 61             node f2;
 62             f2.x=f1.x+dx[i];
 63             f2.y=f1.y+dy[i];
 64             f2.t=f1.t;
 65             if(0<=f2.x&&f2.x<n&&0<=f2.y&&f2.y<m&&!vis1[f2.x][f2.y]&&mp[f2.x][f2.y]!='#'){
 66                 f2.t++;
 67                 if(mp[f2.x][f2.y]=='@'){
 68                     ans1[f2.x][f2.y]=f2.t;
 69                 }
 70                 vis1[f2.x][f2.y]=1;
 71                 q.push(f2);
 72             }
 73         }
 74     }
 75 }
 76 
 77 void bfs2()
 78 {
 79     queue<node> q;
 80     while(!q.empty())
 81         q.pop();
 82     q.push(g2);
 83     while(!q.empty()){
 84         node f1;
 85         f1=q.front();
 86         q.pop();
 87         for(int i=0; i<4; i++){
 88             node f2;
 89             f2.x=f1.x+dx[i];
 90             f2.y=f1.y+dy[i];
 91             f2.t=f1.t;
 92             if(0<=f2.x&&f2.x<n&&0<=f2.y&&f2.y<m&&!vis2[f2.x][f2.y]&&mp[f2.x][f2.y]!='#'){
 93                 f2.t++;
 94                 if(mp[f2.x][f2.y]=='@'){
 95                     ans2[f2.x][f2.y]=f2.t;
 96                 }
 97                 vis2[f2.x][f2.y]=1;
 98                 q.push(f2);
 99             }
100         }    
101     }
102 }
103 
104 int main()
105 {
106     while(~scanf("%d %d",&n,&m)){
107         memset(ans1,-1,sizeof(ans1));
108         memset(ans2,-1,sizeof(ans2));
109         memset(vis1,0,sizeof(vis1));
110         memset(vis2,0,sizeof(vis2));
111         getMp();
112         bfs1();
113         bfs2();
114         int min=INF;
115         for(int i=0;i<n;i++)  
116             for(int j=0;j<m;j++)  
117             {  
118                 if(ans1[i][j]+ans2[i][j]<min&&ans1[i][j]+ans2[i][j]>0)  
119                 {  
120                     min=ans1[i][j]+ans2[i][j];  
121                 }  
122             }
123         printf("%d\n",min*11); 
124     }
125 }

 

posted @ 2016-03-06 15:44  Vmetrio  阅读(271)  评论(0)  编辑  收藏  举报