hdu 1175 连连看（dfs+剪枝）

一些细节没处理好，wa了很多次。

剪枝不强，跑了6000+ms。

View Code

 1 /*

 2 Author:Zhaofa Fang

 3 Lang:C++

 4 */

 5 #include <cstdio>

 6 #include <cstdlib>

 7 #include <sstream>

 8 #include <iostream>

 9 

10 #include <cstring>

11 #include <algorithm>

12 #include <string>

13 #include <utility>

14 #include <vector>

15 #include <queue>

16 #include <stack>

17 #include <map>

18 #include <set>

19 

20 using namespace std;

21 

22 typedef long long ll;

23 #define DEBUG(x) cout<< #x << ':' << x << endl

24 #define PII pair<int,int>

25 #define PB push_back

26 #define MP make_pair

27 #define FI first

28 #define SE second

29 #define lowbit(x) (x&(-x))

30 #define INF (1<<30)

31 

32 const double eps = 1e-6;

33 

34 int maz[1005][1005];

35 int dx[]={1,0,0,-1};

36 int dy[]={0,1,-1,0};

37 bool OK;

38 int n,m;

39 int x1,y1,x2,y2;

40 bool check(int x,int y)

41 {

42     if(x<1 || x>n || y<1 || y>m)return false;

43     return true;

44 }

45 void dfs(int x,int y,int turn,int dire)

46 {

47     if(turn > 2 || !check(x,y))return;

48     if(OK)return ;

49     if(x == x2 && y == y2)

50     {

51         OK = 1;

52         return;

53     }

54     for(int i=0;i<4;i++)

55     {

56         if(dire + i == 3)continue;

57         int xx = x + dx[i];

58         int yy = y + dy[i];

59         if((xx != x2 || yy != y2) && maz[xx][yy] != 0)continue;

60         if(dire != i && dire != -1)dfs(xx,yy,turn + 1,i);

61         else dfs(xx,yy,turn,i);

62     }

63 }

64 

65 int main()

66 {

67     #ifndef ONLINE_JUDGE

68     freopen("in","r",stdin);

69     #endif

70     while(~scanf("%d%d",&n,&m),n,m)

71     {

72         for(int i=1;i<=n;i++)

73         {

74             for(int j=1;j<=m;j++)

75                 scanf("%d",&maz[i][j]);

76         }

77         int q;

78         scanf("%d",&q);

79         while(q--)

80         {

81             scanf("%d%d%d%d",&x1,&y1,&x2,&y2);

82             if(maz[x1][y1] != maz[x2][y2]

83             || maz[x1][y1] == 0 || maz[x2][y2] == 0)

84             {

85                 puts("NO");

86                 continue;

87             }

88             OK=0;

89             dfs(x1,y1,0,-1);

90             if(OK)puts("YES");

91             else puts("NO");

92         }

93     }

94     return 0;

95 }