poj1054 The Troublesome Frog 瞎搞。

鏈接：http://poj.org/problem?id=1054

題意：就是一個格子裏一條線上最長有幾個青蛙（青蛙間隔相同）~。可是其實青蛙的起點重點必須是在外面。

直接寫一個搜就是。

 1 #include <stdio.h>
 2 #include <string.h>
 3 #include <iostream>
 4 #include <algorithm>
 5 #include <stdlib.h>
 6 #include <vector>
 7 #include <queue>
 8 #include <stack>
 9 #define loop(s,i,n) for(i = s;i < n;i++)
10 #define cl(a,b) memset(a,b,sizeof(a))
11 using namespace std;
12 int map[5005][5005];
13 int r,c,n;
14 struct node
15 {
16     int x,y;
17 }p[50005];
18 int cmp( struct node a, struct node b)
19 {
20     if(a.x == b.x)
21     return a.y < b.y;
22     return a.x < b.x;
23 }
24 int judge(int x,int y)
25 {
26     if(x >= 1 && x <= r && y >= 1&&y <= c)
27     return 1;
28 
29     return 0;
30 
31 }
32 int find(struct node tmp,int dx,int dy)
33 {
34 
35     int ans = 2;
36 
37     int leap =1;
38     while(judge(tmp.x+dx,tmp.y+dy))
39     {
40         if(map[tmp.x+dx][tmp.y+dy])
41         {
42             tmp.x+= dx,tmp.y += dy;
43             ans++;
44         }
45         else
46         {
47             leap = 0;
48             break;
49         }
50     }
51     if(leap)
52     return ans;
53     return 0;
54 }
55 int main()
56 {
57     while(~scanf("%d %d",&r,&c))
58     {
59         int i,j;
60         scanf("%d",&n);
61 
62         cl(map,0);
63         for(i = 1;i <= n;i++)
64         scanf("%d %d",&p[i].x,&p[i].y),map[p[i].x][p[i].y] = 1;
65 
66         sort(p+1,p+n+1,cmp);
67         int ans = 0;
68         for(i = 1;i <= n;i++)
69         {
70 
71             for(j = i+1;j <= n;j++)
72             {
73 
74                 int dx,dy;
75                 dx = p[j].x-p[i].x;
76                 dy = p[j].y-p[i].y;
77                 struct node tmp;
78                 tmp = p[j];
79                 if(judge(p[i].x-dx,p[i].y-dy))
80                 continue;
81                 int cnt = find(tmp,dx,dy);
82 
83                 if(ans < cnt)
84                 ans = cnt;
85 
86             }
87         }
88         if(ans < 3)
89         puts("0");
90         else
91         printf("%d\n",ans);
92     }
93     return 0;
94 }

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