BZOJ1270: [BeijingWc2008]雷涛的小猫
1270: [BeijingWc2008]雷涛的小猫
Time Limit: 50 Sec Memory Limit: 162 MBSubmit: 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\n",ans); 84 85 return 0; 86 87 }