Timus 1260. Nudnik photographer Dynamic Planning

Topic Portal

To finish the problem. The DP image is a state in transition, pushed from one (more) current state to the next (many) states.

This topic means that n people want to take pictures, and the Age of N is 1 to n. Standing in a row, the leftmost segment is 1 years old, and the age difference between the two adjacent persons will not exceed 2 years. Ask how many total queuing scenarios there are.

Solution:

The status of individual affiliation can be pushed from the state of I-1. We know that the first person will be at least adjacent to i-1 or i-2. Therefore, [i-1] and [i-2] are used as states. Can be divided into four different states:

①[i-1][i-2] i.e. i-1,i-2 adjacent and i-1 in front of i-2

②[I-2][I-1]

③[i-1][i-2] ....   And [I-2][i-1] .... That is, [I-1][i-2] is adjacent but not at the end (the analysis shows whether I-1,i-2 is close to the next state at the end, and if there is no such state, the state is omitted from the i-1 push to i)

④[i-2] ... [I-1] That is i-1,i-2 non-sequential, the analysis of i-1 must be at the end.

We can quickly conclude that when I am added to the first person, the status ② can be introduced status ①, state ①④ can be launched status ②, the status ①③ can be launched ③,①④ can launch ②.

Thus the state-state transition equation:

DP[1][I]=DP[2][I-1];
DP[2][I]=DP[2][I-1]+DP[4][I-1];
DP[3][I]=DP[1][I-1]+DP[3][I-1];
DP[4][I]=DP[1][I-1];

Code

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespacestd;intMain () {intdp[5][ -]; Memset (DP,0,sizeof(DP)); dp[1][1]=1; dp[2][2]=1;  for(intI=3; i<= -; i++) {dp[1][i]=dp[2][i-1]; dp[2][i]=dp[2][i-1]+dp[4][i-1]; dp[3][i]=dp[1][i-1]+dp[3][i-1]; dp[4][i]=dp[1][i-1]; }    intN;  while(~SCANF ("%d",&N)) {cout<<dp[1][n]+dp[2][n]+dp[3][n]+dp[4][n]<<Endl; }    return 0;}

Later, it feels simpler to have another code on the Web.

The seat is DP, when the first is 1, the second is 2, the combination is dp[i-1]; When the first one is 1, the second is 3, the third one is 2, the combination is dp[i-3]; and the last case is 1357 ... 8642.

So the DP equation is dp[i] = dp[i-1]+dp[i-3]+1.

#include <stdio.h>intN;intdp[ -];intMain () {scanf ("%d", &N); dp[1] =1; dp[2] =1; dp[3] =2; dp[4] =4;  for(inti =5; I <= -; ++i) {Dp[i]= dp[i-1]+dp[i-3]+1; } printf ("%d", Dp[n]); return 0;}

Timus 1260. Nudnik photographer Dynamic Planning