HDU 1026 Ignatius and the Princess I(BFS）

题目链接

复杂啊，这到底是模拟题还是BFS啊。。。。数组开小RE一次，2Y。。。调试了好久啊。

  1 #include <stdio.h>
  2 #include <string.h>
  3 #define N 1000001
  4 char p[101][101];
  5 int o[101][101],key[101][101];// key标记,o标记
  6 int r[100001],c[100001],ti[100001];//ti记录时间
  7 int past[100001],load[100001];//记录路径
  8 int main()
  9 {
 10     int n,m,i,j,k,start,end,num,z;
 11     int a[4] = {0,0,1,-1};
 12     int b[4] = {1,-1,0,0};
 13     while(scanf("%d%d%*c",&n,&m)!=EOF)
 14     {
 15         memset(o,0,sizeof(o));
 16         memset(key,0,sizeof(key));
 17         for(i = 0; i <= n-1; i ++)
 18             gets(p[i]);
 19         if(p[n-1][m-2] == 'X'&&p[n-2][m-1] == 'X')//小剪枝
 20         {
 21             printf("God please help our poor hero.\n");
 22             printf("FINISH\n");
 23             continue;
 24         }
 25         for(i = 0; i <= n-1; i ++)
 26             for(j = 0; j <= m-1; j ++)
 27             {
 28                 if(p[i][j] == 'X')
 29                 {
 30                     o[i][j] = N;
 31                     key[i][j] = 1;
 32                 }
 33                 else if(p[i][j] == '.')
 34                     o[i][j] = 0;
 35                 else if(p[i][j] >= '0'&&p[i][j] <= '9')
 36                     o[i][j] = p[i][j] - '0';
 37             }
 38         start = end = 1;
 39         r[1] = c[1] = 0;
 40         ti[1] = o[0][0];
 41         num = 0;
 42         z = 0;
 43         key[0][0] = 1;
 44         while(start <= end)//BFS
 45         {
 46             j = 1;
 47             if(z) break;
 48             for(i = start; i <= end&&!z; i ++)
 49             {
 50                 if(ti[i] == 0)
 51                 {
 52                     for(k = 0; k <= 3&&!z; k ++)
 53                     {
 54                         if(r[i]+a[k]>=0&&c[i]+b[k]>=0&&r[i]+a[k]<=n-1&&c[i]+b[k]<=m-1)
 55                         {
 56                             if(key[r[i]+a[k]][c[i]+b[k]] == 0)
 57                             {
 58                                 r[j + end] = r[i]+a[k];
 59                                 c[j + end] = c[i]+b[k];
 60                                 ti[j + end] = o[r[i]+a[k]][c[i]+b[k]];
 61                                 key[r[i]+a[k]][c[i]+b[k]] = 1;
 62                                 past[j+end] = i;
 63                                 if(r[j + end] == n-1&&c[end + j] == m-1&&ti[j+end] == 0)
 64                                 {
 65                                     z = 1;
 66                                 }
 67                                 j ++;
 68                             }
 69                         }
 70                     }
 71                 }
 72                 else
 73                 {
 74                     r[j + end] = r[i];
 75                     c[j + end] = c[i];
 76                     ti[j + end] = ti[i] - 1;
 77                     past[j+end] = i;
 78                     if(r[j + end] == n-1&&c[end + j] == m-1&&ti[j+end] == 0)
 79                     {
 80                         z = 1;
 81                     }
 82                     j ++;
 83                 }
 84             }
 85             start = end + 1;
 86             end = end + j - 1;
 87             num ++;
 88         }
 89         if(z)
 90         {
 91             printf("It takes %d seconds to reach the target position, let me show you the way.\n",num);
 92             load[0] = end;//找路径
 93             for(i = 1;;i ++)
 94             {
 95                if(past[load[i-1]] == 1)
 96                {
 97                    load[i] = 1;
 98                    break;
 99                }
100                else
101                load[i] = past[load[i-1]];
102             }
103             for(j = i;j >= 1;j --)
104             {
105                 printf("%ds:",i-j+1);
106                 if(r[load[j]] == r[load[j-1]]&&c[load[j]] == c[load[j-1]])
107                 printf("FIGHT AT (%d,%d)\n",r[load[j]],c[load[j]]);
108                 else
109                 printf("(%d,%d)->(%d,%d)\n",r[load[j]],c[load[j]],r[load[j-1]],c[load[j-1]]);
110             }
111             printf("FINISH\n");
112         }
113         else
114             printf("God please help our poor hero.\nFINISH\n");
115     }
116     return 0;
117 }
posted @ 2012-07-09 13:35  Naix_x  阅读(146)  评论(0)    收藏  举报