USACO 2.4.1 The Tamworth Two 题解

【算法】模拟　　【难度】★☆☆☆☆

直接模拟。
注意如果牛和人开始绕圈子会陷入死循环，所以在时间足够大的时候退出，具体的临界值我选了10000000，nocow上有证明说最多160000步。

【收获】极限情况判断

View Code 
  1 /*
  2 ID: wsc5001
  3 LANG: C
  4 TASK: ttwo
  5 */
  6 #include <stdio.h>
  7 #include <stdlib.h>
  8 struct wz{int x,y,fs;};
  9 int ct=0;
 10 int map[12][12];// 0=可通过√ 1=墙x
 11 struct wz moveit(struct wz c)
 12 {
 13    struct wz ans;
 14    if (c.fs==0)//↑
 15    {
 16        if (map[c.x-1][c.y]==1||map[c.x-1][c.y]==-1)
 17        {
 18           ans=c;
 19           ans.fs=1;
 20           return ans;
 21        }
 22        else
 23        {
 24            ans=c;
 25            ans.x=c.x-1;
 26            return ans;
 27        }
 28    }
 29    if (c.fs==1)//→
 30    {
 31        if (map[c.x][c.y+1]==1||map[c.x][c.y+1]==-1)
 32        {
 33           ans=c;
 34           ans.fs=2;
 35           return ans;
 36        }
 37        else
 38        {
 39            ans=c;
 40            ans.y=c.y+1;
 41            return ans;
 42        }
 43    }
 44    if (c.fs==2)//↓
 45    {
 46        if (map[c.x+1][c.y]==1||map[c.x+1][c.y]==-1)
 47        {
 48           ans=c;
 49           ans.fs=3;
 50           return ans;
 51        }
 52        else
 53        {
 54            ans=c;
 55            ans.x=c.x+1;
 56            return ans;
 57        }
 58    }
 59    if (c.fs==3)//←
 60    {
 61        if (map[c.x][c.y-1]==1||map[c.x][c.y-1]==-1)
 62        {
 63           ans=c;
 64           ans.fs=0;
 65           return ans;
 66        }
 67        else
 68        {
 69            ans=c;
 70            ans.y=c.y-1;
 71            return ans;
 72        }
 73    }
 74 }
 75 int main()
 76 {
 77    freopen("ttwo.in","r",stdin);
 78    freopen("ttwo.out","w",stdout);
 79    int i,j,tms=0;
 80    char temp;
 81    struct wz f,c,sf,sc;
 82    for (i=0;i<12;i++)
 83        for (j=0;j<12;j++)
 84            map[i][j]=-1;
 85    for (i=1;i<=10;i++)
 86    {
 87        for (j=1;j<=10;j++)
 88        {
 89            scanf("%c",&temp);
 90            if(temp=='.')   map[i][j]=0;
 91            if(temp=='*')   map[i][j]=1;
 92            if(temp=='C')   {map[i][j]=0;c.x=i;c.y=j;}
 93            if(temp=='F')   {map[i][j]=0;f.x=i;f.y=j;}
 94        }
 95        scanf("%c",&temp);
 96    }
 97    sf.x=f.x;   sf.y=f.y;
 98    sc.x=c.x;   sc.y=c.y;
 99    
100    c.fs=0; f.fs=0;
101    
102    while (ct<=10000000)
103    {
104        c=moveit(c);
105        f=moveit(f);
106        ct++;
107        if (c.x==f.x && c.y==f.y)
108            {printf("%d\n",ct);break;}
109        if (c.x==sc.x && c.y==sc.y && f.x==sf.x && f.y==sf.y)
110            tms++;
111        if (tms==4)
112            {printf("0\n");break;}
113    }
114    if (ct>10000000)printf("0\n");
115    //system("pause");
116    fclose(stdin);
117    fclose(stdout);
118 }



posted @ 2012-03-25 19:40  wsc500  阅读(183)  评论(0)  编辑  收藏  举报