Hdu2084 _ shuita [simple question]

Shuta

Time Limit: 1000/1000 MS (Java/others) memory limit: 32768/32768 K (Java/others) total submission (s): 22488 accepted submission (s): 13555
Problem description
When talking about the DP algorithm, a classic example is the data tower problem, which is described as follows:

There is a number tower as shown below. It is required to go from the top layer to the bottom layer. If each step can only go to adjacent nodes, what is the maximum sum of the numbers of the nodes that pass through?
I already told you that this is a DP question. Can you AC it?
 
Input
The input data first includes an integer c, indicating the number of test instances. The first row of each test instance is an integer N (1 <= n <= 100 ), the height of the tower. Next, use N rows of numbers to represent the tower. Row I has an I integer, and All integers are within the range [0, 99.
 
Output
For each test instance, the output is the largest possible sum, and each instance occupies one row of output.
 
Sample Input
1
5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
 
Sample output
30
 
Source

/1/15 ACM Program Design Final Examination

Good understanding

Idea: If from above consideration, each time there are two choices, N layer tower has 2 ^ (N-1) A solution,

The traversal side is unscientific. In another way, consider from the bottom up, each time compare the size of near two numbers, let

Add a large number, so that each selection is the best case, move up and accumulate layer by layer, and finally add

At the top, the result is optimal, that is, the maximum.

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int map[110][110],dp[110];int main(){    int C,N;    scanf("%d",&C);    while(C--)    {        scanf("%d",&N);        memset(map,0,sizeof(map));        memset(dp,0,sizeof(dp));        for(int i = 1; i <= N;i++)        {            for(int j = 1; j <= i; j++)            {                scanf("%d",&map[i][j]);            }        }        for(int i = N; i>=1; i--)        {            for(int j = 1; j <= i;j++)            {                if(i == N)                    dp[j] = map[i][j];                else                {                    dp[j] = max(dp[j],dp[j+1]) + map[i][j];                }            }        }        printf("%d\n",dp[1]);    }    return 0;}

Hdu2084 _ shuita [simple question]