Sequence Transformation (LIS variant)

Sequence transformations

Time limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others) total submission (s): Accepted Submission (s): 448

Problem description We have a series of a1,a2 ... An, you now want to modify the least number of elements, so that the sequence is strictly incremented. Each element must be an integer, either before or after the modification. Please output the minimum number of elements you need to modify.

Input first line Enter aT(1≤t≤) That indicates how many sets of data
Each group of data:
The first line enters an(1≤n≤5) , which indicates the length of the sequence
Enter n number in the second linea1,a2,.. . ,An 。
Each element in the sequence is a positive integer and does not exceed Ten6.

Output for each set of data, first a row
Case #i:
Then output the minimum number of elements that need to be modified.

Sample Input221 1032 5 4

Sample outputcase #1:0Case #2:1

SOURCE2015 year Baidu Star Program Design Competition-Preliminary (2)

Exercises

Changing series makes the monotone increment sub-sequence; then it just needs to be dp[i]-dp[j]>=i-j, and the method of monotone increment subsequence is added;

Dp[i]-dp[j]>=i-j is the dp[i]-i>=dp[j]-j;

Code:

#include <iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#include<vector>using namespacestd;Const intMAXN = 1e5 + -;intNUM[MAXN];intDp[maxn];vector<int>Vec;/*int main () {int T, N, Kase = 0;    scanf ("%d", &t);        while (t--) {vec.clear ();        scanf ("%d", &n);            for (int i = 0; i < N; i++) {scanf ("%d", num + i);            Num[i]-= i;            if (Upper_bound (Vec.begin (), Vec.end (), num[i]) = = Vec.end ()) {Vec.push_back (num[i]);                } else{int p = upper_bound (Vec.begin (), Vec.end (), num[i])-Vec.begin ();            VEC[P] = Num[i];    }} printf ("Case #%d:\n%d\n", ++kase, N-vec.size ()); } return 0;}*/intMain () {intT, N, Kase =0; scanf ("%d", &T);  while(t--) {scanf ("%d", &N); Memset (DP,0,sizeof(DP)); intAns =0;  for(inti =0; i < N; i++) {scanf ("%d", Num +i);  for(intj =0; J < I; J + +){                if(Num[i]-num[j] >= i-j) {Dp[i]= Dp[j] +1; Ans=Max (ans, dp[i]); }}} printf ("Case #%d:\n%d\n", ++kase, N-ans-1); }    return 0;}

Sequence Transformation (LIS variant)