Test Instructions:N-layer bricks are placed in a groove, the top layer has n blocks, the second layer has n-1 blocks, ..., and the bottom layer has only one brick. The bricks on layer I are numbered from left to right and 1,2,......,i, and block J of Layer I has a value a[i,j] (a[i,j]<=50). Here is an example of a 5-storey brick:
If you want to knock off the first layer of block J brick, if i=1, you can directly knock it off, if i>1, you must first knock off the i-1 layer of the section J and j+1 block bricks. Your task is to knock off (m<=500) bricks from a heap of bricks with N (n<=50) so that the value of the bricks knocked out is the largest sum.
Method:Dynamic planning
parsing:Ignore the 76 value in the template diagram. Test this problem, the first 1 hours of basic ratio, the feeling is to use dynamic planning but did not find the right method. Then began to put the brain hole, began to rotate the picture, found to rotate to a magical figure, the problem is changed? For example, let's look at a sample
So we can not find any problems, so we have to change the angle.
We can find that the premise of taking 8 is to take a line of 2 and go to its front 2 to take the third line to the far right of the 2 premise is to take the line all and the top row, and all the elements in front of him. And so on, a law obviously appears, we may wish to set f[i][j][k] to indicate that line I take the former J and take a total of k points. The reason why this is set is to take a line must be a prefix. How to update it? is from the previous row fetch P update, wherein p>=j-1, this is obvious, because if the current line you want to fetch to the first J, then the last line of the j-1 must be taken, but also to take the j-1 before the prefix need to take, the equation of the motion of the column. One thing to note, because the enumeration J may be zero, that is, p may be-1, here you need to judge, (because here hang 5 points good pit!) )
Code:
#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define N -using namespace STD;intA[n][n];intSum[n][n];intf[n][n][n*Ten];intN,m;intMain () {scanf("%d%d", &n,&m); for(intj=1; j<=n;j++) { for(inti=j;i<=n;i++) {scanf("%d", &a[i][j]); } } for(intI=1; i<=n;i++) for(intj=1; j<=i;j++) sum[i][j]=sum[i][j-1]+A[I][J];intans=-1;memset(f,-1,sizeof(f)); for(intI=1; i<=n;i++) f[i][0][0]=0, f[i][1][1]=a[i][1]; for(intI=1; i<=n;i++) { for(intj=0; j<=i;j++) { for(intk=0; k<=m;k++) {if(j<=k) { for(intp=j-1>=0? J-1:0;p <=i-1;p + +) {if(f[i-1][p][k-j]!=-1) {F[i][j][k]=max (f[i][j][k],f[i-1][P][K-J]+SUM[I][J]); } } }if(k==m) Ans=max (Ans,f[i][j][k]); } } }printf("%d\n", ans);}
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Hnoi 2004 Brick Dynamic Planning
