FateTime
limit:2000/1000 MS (java/others) Memory limit:32768/32768 K (java/others)
Total submission (s): 12611 Accepted Submission (s): 4401
Problem description through the valley means that the lemon of the great Lord has been infinitely close!
But who can think, Yifenfei in kill some cats, but again face the fate of the maze of the test, this is the Devil lemon set up another organ. You know, whoever, if trapped in the maze for more than 1 hours, will surely die!
Poor Yifenfei in order to save mm, righteousness without return to jump into the maze. Let us help to help the persistent him!
The great maze of destiny can be seen as a two-dimensional grid array, as shown in:
Yifenfei first in the upper left corner, the aim of course is to reach the lower right corner of the great Lord's seat. Every lattice in the maze is cursed by the goddess of luck or the devil, so each grid corresponds to a value, where it automatically gets the corresponding value.
It is now stipulated that Yifenfei can only go to the right or down and only one grid at a time. But if you go to the right, you can walk one block at a time or go to the number of columns that are currently in the number of columns, that is: if the current lattice is (x, y), the next step can be (x+1,y), (x,y+1) or (x,y*k) where k>1.
In order to be able to best grasp the eradication of the Devil Lemon,yifenfei hope to be able to get the greatest fortune in this maze of fate.
Input data is first an integer c, which represents the number of groups of test data.
The first line of each set of test data is two integer n,m, representing the number of rows and columns (1<=n<=20,10<=m<=1000);
followed by n rows of data, each containing M integers representing the lucky value K (|k|<100) for the lattice of the N row M column.
Output match each set of test data with an integer that represents the maximum lucky value that Yifenfei can get.
Sample Input
13 89 10 10 10 10-10 10 1010-11-1 0 2 11 10-20-11-11 10 11 2 10-10-10
Sample Output
52
Authoryifenfei
SOURCEACM Program Design Final Exam 081230
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#include <stdio.h> #include <string.h> #include <algorithm>using namespace Std;const int inf=-1< <30;int A[25][1010],dp[25][1010];int Main () {int c;scanf ("%d", &c), while (c--) {int n,m,i,j,k,maxn;scanf ("%d%d") , &n,&m); for (I=1;i<=n;++i) {for (j=1;j<=m;++j) {scanf ("%d", &a[i][j]);}} Memset (Dp,0,sizeof (DP)); for (j=0;j<=m;++j) for (i=0;i<=n;++i) dp[i][j]=inf;dp[1][1]=a[1][1];for (i=1;i<=n; ++i) {for (j=1;j<=m;++j) { Dp[i+1][j]=max (dp[i+1][j],dp[i][j]+a[i+1][j]); Dp[i][j+1]=max (dp[i][j+1],dp[i][j]+a[i][j+1]); for (k=2;k*j<=m;++k) { Dp[i][k*j]=max (dp[i][k*j],dp[i][j]+a[i][k*j]);}}} printf ("%d\n", Dp[n][m]);} return 0;}
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hdoj-2571-Destiny "Dynamic Planning"
