BZOJ1270: [BeijingWc2008]雷涛的小猫

1270: [BeijingWc2008]雷涛的小猫

Time Limit: 50 Sec  Memory Limit: 162 MB
Submit: 1338  Solved: 707
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Description

Input

Output

Sample Input

Sample Output

8

HINT

Source

 

dp水题

题解见代码注释

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <vector>
 8 #define min(a, b) ((a) < (b) ? (a) : (b))
 9 #define max(a, b) ((a) > (b) ? (a) : (b))
10 #define abs(a) ((a) < 0 ? (-1 * (a)) : (a))
11 inline void swap(int &a, int &b)
12 {
13     int tmp = a;a = b;b = tmp;
14 }
15 inline void read(int &x)
16 {
17     x = 0;char ch = getchar(), c = ch;
18     while(ch < '0' || ch > '9') c = ch, ch = getchar();
19     while(ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
20     if(c == '-') x = -x;
21 }
22 
23 const int INF = 0x3f3f3f3f;
24 const int MAXN = 5000 + 10;
25 
26 //dp[i][j]表示当前高度为i，在第j颗树上的最大数量 
27 //ma[i]表示在高度为i时的最大数量
28 //dp[i][j] = max(dp[i - 1][j], ma[i - delta]) + val[i][j]
29 //ma[i] = max(ma[i], ma[j])
30 
31 int dp[MAXN], ma[MAXN], val[MAXN][MAXN], n, h, delta;
32 
33 int main()
34 {
35     read(n), read(h), read(delta);
36     for(register int i = 1;i <= n;++ i)
37     {
38         int tmp;read(tmp);
39         for(register int j = 1;j <= tmp;++ j)
40         {
41             int tt;
42             read(tt);
43             ++ val[tt][i];
44         }
45     }
46     int ans = 0;
47     for(register int i = 1;i <= h;++ i)
48         for(register int j = n;j >= 1;-- j)
49         {
50             if(i - delta >= 0) dp[j] = max(dp[j], ma[i - delta]);
51             dp[j] += val[i][j];
52             ma[i] = max(ma[i], dp[j]);
53             ans = max(ans, dp[j]);
54         }
55     printf("%d", ans);
56     return 0;
57 }
BZOJ1270

 

posted @ 2017-12-04 23:29  嘒彼小星  阅读(178)  评论(0)  编辑  收藏  举报