poj1054The Troublesome Frog

链接

想O（n*n)的DP  怎么想都超内存 看讨论有说hash+DP过的 实现比较繁琐

大部分直接暴力过了 

直接枚举每个i j 与他们在一条线上的点 是不是给出的点 

注意它必须能跳进和跳出 

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 using namespace std;

 7 #define N 5010

 8 bool flag[N][N];

 9 struct node

10 {

11     int x,y;

12 }p[N];

13 bool cmp(node a,node b)

14 {

15     if(a.x==b.x)

16     return a.y<b.y;

17     return a.x<b.x;

18 }

19 int main()

20 {

21     int i,j,k,n,r,c;

22     while(scanf("%d%d",&r,&c)!=EOF)

23     {

24         memset(flag,0,sizeof(flag));

25         scanf("%d",&n);

26         for(i = 1; i <= n ;i++)

27         {

28             scanf("%d%d",&p[i].x,&p[i].y);

29             flag[p[i].x][p[i].y] = 1;

30         }

31         sort(p+1,p+n+1,cmp);

32         int ans=0,ff=1;

33         for(i = 1 ; i <= n ; i++)

34             for(j = i+1; j <= n ; j++)

35             {

36                 int tx = p[j].x-p[i].x;

37                 int ty = p[j].y - p[i].y;

38                 int ox = p[j].x;

39                 int oy = p[j].y;

40                 if(p[i].x-tx>=1&&p[i].x-tx<=r&&p[i].y-ty>=1&&p[i].y-ty<=c)

41                 continue;

42                 ff = 1;

43                 for(k = 2 ; ; k++)

44                 {

45                     ox+=tx;

46                     oy+=ty;

47                     if(ox>r||oy>c||ox<1||oy<1)

48                     break;

49                     if(!flag[ox][oy])

50                     {

51                         ff = 0;

52                         break;

53                     }

54                 }

55                 if(ff)

56                 ans = max(ans,k);

57             }

58         if(ans>2)

59         printf("%d\n",ans);

60         else

61         printf("0\n");

62     }

63     return 0;

64 }

View Code