BZOJ 4993: [Usaco2017 Feb]Why Did the Cow Cross the Road II LCS

4993: [Usaco2017 Feb]Why Did the Cow Cross the Road II

Time Limit: 10 Sec  Memory Limit: 256 MB
Submit: 28  Solved: 19
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Description

上下有两个长度为n、位置对应的序列A、B，

其中数的范围均为1~n。若abs(A[i]-B[j])<= 4，则A[i]与B[j]间可以连一条边。

现要求在边与边不相交的情况下的最大的连边数量。

n <= 10^3

Input

Output

Sample Input

6
1
2
3
4
5
6
6
5
4
3
2
1

Sample Output

5

裸LCS

刷水有益健康

#include<cmath>#include<ctime>#include<cstdio>#include<cstring>#include<cstdlib>#include<iostream>#include<algorithm>#include<iomanip>#include<vector>#include<string>#include<bitset>#include<queue>#include<set>#include<map>using namespace std;typedef long long ll;inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}return x*f;}void print(int x){if(x<0)x=-x,putchar('-');if(x>=10)print(x/10);putchar(x%10+'0');}int n,x[1007],y[1007],dp[1007][1007];int main() {        n=read();for(int i=1 ; i<=n ; i++) x[i]=read();for(int i=1 ; i<=n ; i++) y[i]=read();for(int i=1 ; i<=n ; i++){for(int j=1 ; j<=n ; j++){if(abs(x[i]-y[j]) <= 4) dp[i][j] = dp[i-1][j-1]+1;else dp[i][j] = max(dp[i][j-1],dp[i-1][j]);}}cout << dp[n][n] << '\n';    return 0;}