HDU 2084 number Tower (DP)

Number of towersTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d &%i64 U SubmitStatusPracticeHDU 2084

Description

When it comes to the DP algorithm, a classic example is the problem of the tower, which is described in this way:

As shown in the following tower, required to go from the top floor to the bottom, if each step can only go to adjacent nodes, then the number of nodes through the sum of the maximum is how much?

Already told you, this is a DP problem, can you AC?

Input

The input data first consists of an integer c, which indicates the number of test instances, the first line of each test instance is an integer N (1 <= N <= 100), which represents the height of the tower, followed by the number of columns in n rows, where line I has an integer i, and all integers are within the interval [0,99].

Output

For each test instance, the output may be the largest and one row per instance output.

Sample Input

1573 88 1 0 2 7 4 44 5 2 6 5

Sample Output

30

Idea: From the bottom up, each position of the DP element is the maximum value of this vertex, the first line of the DP value is equal to itself, and then each rise one line to fill out the top element of the DP value. (Interesting is the C + + compiled after tle, instead of g++ to become the fastest solution, laugh and cry)

1#include <stdio.h>2#include <stdlib.h>3#include <string.h>4 #defineMAX 1055 6 intMainvoid)7 {8     intC,n;9     intDp[max][max];Ten  Onescanf"%d",&c); A      while(C--) -     { -scanf"%d",&n); the          for(inti =1; I <= N;i + +) -              for(intj =1; J <= I;j + +) -scanf"%d",&dp[i][j]); -  +          for(inti = n-1; I >=1; I--) -              for(intj =1; J <= I;j + +) +DP[I][J] + = Dp[i +1][J] > Dp[i +1][j +1] ? Dp[i +1][J]: Dp[i +1][j +1]; Aprintf"%d\n", dp[1][1]); at     } -  -     return    0; -}

HDU 2084 number Tower (DP)