Dfs+dp
DFS enumeration for each case, each layer recursively determines the number of k i:i = a[k-1]+1 to A[k-1]*n+1
When you have finished enumerating a sequence, use check () to test the max that it can achieve
Use DP. Set Dp[i] to the minimum number of the face value of I
for (int k=1; a[k]<=i&&k<=t;k++) { if(dp[i-a[k]]<N) { Dp[i]=min (dp[i], dp[i-a[k]]+1);} }
Note Dp[i] Initially set to maximum value
The code is as follows:
#include <iostream>#include<cstring>#defineSize 41using namespacestd;intA[size];intdp[10005];intBest[size];intN,t;//N-Sheet t typeintnum=0;voidSaveintx) {num=x; for(intI=1; i<=t;i++) {Best[i]=A[i]; }}intCheck () {//memset (Dp,0x3f,sizeof (DP));dp[0]=0; intans=0; BOOLok=true; //for (int i=1;i<=t;i++) cout<<a[i]<< ""; //cout<<endl; for(intI=1; ok;i++) {OK=false; Dp[i]=0x3f3f3f3f; for(intk=1; a[k]<=i&&k<=t;k++){ if(dp[i-a[k]]<N) {Dp[i]=min (Dp[i], dp[i-a[k]]+1); OK=true; } } if(OK) ans=i; //cout<<dp[i]<< "; } //cout<<endl<<endl; returnans;}voidDfsintk) {if(k>t) { intx=check (); if(x>num) Save (x); return; } for(inti=a[k-1]+1; i<=n*a[k-1]+1; i++) {A[k]=i; DFS (k+1); }}intMain () {CIN>>n>>T; a[1]=1; Num=1; DFS (2); for(intI=1; i<=t;i++) {cout<<best[i]<<' '; } cout<<endl<<"max="<<num<<Endl; cout<<F; return 0;}
By the way:a[1] must be 1!
PS: You can see that I commented out the first line of the check memset, because most of the time, many of the elements behind the DP array are useless (max is not that big), so you just need to write a dp[i]=x03f3f3f3f in the loop. Memset makes the program slow and slow. (It's really slow ...) (You really don't need to try it too slow, because I've spent one hours trying it for you countless times ...) )
If you want to continue optimization, you can maintain the values of the DP and current MAX2 on the DFS side so that the DFS loop can be written like this:
i = a[k-1]+1 to Max2+1
If i>max2+1, then max2+1 this number will always be unable to make up, the result will not be more excellent
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