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 }