CodeForces - 582B Once Again... （LIS变型）好题

CodeForces - 582B Once Again... Time Limit: 1000MS   Memory Limit: 262144KB   64bit IO Format: %I64d & %I64u Submit Status Description You are given an array of positive integers a1, a2, ..., an × T of length n × T. We know that for any i > n it is true that ai = ai - n. Find the length of the longest non-decreasing sequence of the given array. Input The first line contains two space-separated integers: n, T (1 ≤ n ≤ 100, 1 ≤ T ≤ 107). The second line contains n space-separated integers a1, a2, ..., an (1 ≤ ai ≤ 300). Output Print a single number — the length of a sought sequence. Sample Input Input 4 3 3 1 4 2 Output 5 Hint The array given in the sample looks like that: 3, 1, 4, 2, 3, 1, 4, 2, 3, 1, 4, 2. The elements in bold form the largest non-decreasing subsequence. Source Codeforces Round #323 (Div. 1) //题意：输入一个n，T,接下来输入n个数 表示将这含有n个数的序列循环T次后，问这个新的序列最长的不下降的序列为多长。 //思路： 因为n很小所以可以先求出循环n次的序列的最长不下降序列，然后再拓展到循环T次的最长长度。 #include<cstdio> #include<cmath> #include<cstring> #include<algorithm> #define N 10010 using namespace std; int dp1[N]={0},dp2[N]={0},dp3[310]={0}; int a[N]; int main() { int n,m,i,j,p; while(scanf("%d%d",&n,&m)!=EOF) { memset(dp3,0,sizeof(dp3)); for(i=1;i<=n;i++) { scanf("%d",&a[i]); dp3[a[i]]++; } for(i=1;i<=n*n;i++) a[i]=a[(i-1)%n+1]; int l=min(n,m); l*=n; for(i=1;i<=l;i++) { int tmp=0; for(j=0;j<i;j++) if(a[j]<=a[i]) tmp=max(tmp,dp1[j]); dp1[i]=tmp+1; } for(i=l;i>=1;i--) { int tmp=0; for(j=l+1;j>i;j--) if(a[j]>=a[i]) tmp=max(tmp,dp2[j]); dp2[i]=tmp+1; } int ans=0; if(m<=n) { for(i=1;i<=l;i++) ans=max(ans,dp1[i]+dp2[i]-1); } else { for(i=1;i<=l;i++) ans=max(ans,dp1[i]+dp2[i]-1+dp3[a[i]]*(m-n)); } printf("%d\n",ans); } return 0; }