BZOJ1270: [BeijingWc2008]雷涛的小猫

1270: [BeijingWc2008]雷涛的小猫

Time Limit: 50 Sec  Memory Limit: 162 MB
Submit: 836  Solved: 392
[Submit][Status]

Description

Input

Output

Sample Input

Sample Output

8

HINT

Source

 题解:

水水的DP。。。

考虑到n*m很小,所以我们可以从下往上DP,记录一个该行的最大值就可以O(1)转移了。

代码:

 1 #include<cstdio>
 2 
 3 #include<cstdlib>
 4 
 5 #include<cmath>
 6 
 7 #include<cstring>
 8 
 9 #include<algorithm>
10 
11 #include<iostream>
12 
13 #include<vector>
14 
15 #include<map>
16 
17 #include<set>
18 
19 #include<queue>
20 
21 #include<string>
22 
23 #define inf 1000000000
24 
25 #define maxn 2100
26 
27 #define maxm 500+100
28 
29 #define eps 1e-10
30 
31 #define ll long long
32 
33 #define pa pair<int,int>
34 
35 #define for0(i,n) for(int i=0;i<=(n);i++)
36 
37 #define for1(i,n) for(int i=1;i<=(n);i++)
38 
39 #define for2(i,x,y) for(int i=(x);i<=(y);i++)
40 
41 #define for3(i,x,y) for(int i=(x);i>=(y);i--)
42 
43 #define mod 1000000007
44 
45 using namespace std;
46 
47 inline int read()
48 
49 {
50 
51     int x=0,f=1;char ch=getchar();
52 
53     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
54 
55     while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
56 
57     return x*f;
58 
59 }
60 int f[maxn][maxn],n,m,h,mx[maxn];
61 
62 int main()
63 
64 {
65 
66     freopen("input.txt","r",stdin);
67 
68     freopen("output.txt","w",stdout);
69 
70     m=read();n=read();h=read();
71     for1(i,m)
72      {
73          int x=read();
74          for1(j,x)f[i][read()]++;
75      }
76     for1(i,n)
77      {
78         for1(j,m)f[j][i]+=max(f[j][i-1],mx[max(i-h,0)]);
79         for1(j,m)mx[i]=max(mx[i],f[j][i]);
80      }
81     int ans=0; 
82     for1(i,m)ans=max(ans,f[i][n]);
83     printf("%d
",ans); 
84 
85     return 0;
86 
87 }
View Code
原文地址:https://www.cnblogs.com/zyfzyf/p/3979064.html