标签：esc ble first fir mda math efi getc tchar
Gargari got bored to play with the bishops and now, after solving the problem about them, he is trying to do math homework. In a math book he have found k permutations. Each of them consists of numbers 1, 2, ..., n in some order. Now he should find the length of the longest common subsequence of these permutations. Can you help Gargari?
You can read about longest common subsequence there: https://en.wikipedia.org/wiki/Longest_common_subsequence_problem
Input
The first line contains two integers n and k (1 ≤ n ≤ 1000; 2 ≤ k ≤ 5). Each of the next k lines contains integers 1, 2, ..., n in some order — description of the current permutation.
Output
Print the length of the longest common subsequence.
Examples
4 3
1 4 2 3
4 1 2 3
1 2 4 3
3
Note
The answer for the first test sample is subsequence [1, 2, 3].
sol:求m个序列的lcs，似乎看上去很难得样子，实际上想到dp状态就不难了，dp[i]表示以数字 i 结尾的m个序列的lcs长度，然后判断一下是否所有数字 j 都在 i 之前就可以转移了
#include <bits/stdc++.h> using namespace std; typedef int ll; inline ll read() { ll s=0; bool f=0; char ch=‘ ‘; while(!isdigit(ch)) { f|=(ch==‘-‘); ch=getchar(); } while(isdigit(ch)) { s=(s<<3)+(s<<1)+(ch^48); ch=getchar(); } return (f)?(-s):(s); } #define R(x) x=read() inline void write(ll x) { if(x<0) { putchar(‘-‘); x=-x; } if(x<10) { putchar(x+‘0‘); return; } write(x/10); putchar((x%10)+‘0‘); return; } #define W(x) write(x),putchar(‘ ‘) #define Wl(x) write(x),putchar(‘\n‘) const int N=1005; int n,m,a[10][N],Pos[10][N]; int dp[N]; int main() { int i,j,k; R(n); R(m); for(i=1;i<=m;i++) { for(j=1;j<=n;j++) { R(a[i][j]); Pos[i][a[i][j]]=j; } } dp[a[1][1]]=1; for(i=2;i<=n;i++) { for(j=1;j<i;j++) { bool Flag=1; for(k=2;k<=m&&Flag;k++) if(Pos[k][a[1][j]]>Pos[k][a[1][i]]) Flag=0; if(Flag) dp[a[1][i]]=max(dp[a[1][i]],dp[a[1][j]]+1); } if(!dp[a[1][i]]) dp[a[1][i]]=1; } int ans=1; for(i=1;i<=n;i++) ans=max(ans,dp[i]); Wl(ans); return 0; } /* Input 4 3 1 4 2 3 4 1 2 3 1 2 4 3 Output 3 */
标签：esc ble first fir mda math efi getc tchar
原文地址：https://www.cnblogs.com/gaojunonly1/p/10764300.html
