FZU 1004 Number Triangle(简单DP)

Consider the number triangle shown below. Write a program that calculates the highest sum of numbers that can be passed on a route that starts at the top and ends somewhere on the base. Each step can go either diagonally down to the left or diagonally down to the right.

    7  3   88   1   02   7   4   44   5   2   6   5

In the sample above, the route from 7 to 3 to 8 to 7 to 5 produces the highest sum: 30.

Input

There are multiple test cases.The first line of each test case contains R (1 <= R <= 1000), the number of rows. Each subsequent line contains the integers for that particular row of the triangle. All the supplied integers are non-negative and no larger than 100.
Output

Print a single line containing the largest sum using the traversal specified for each test case.
Sample Input

5
7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
Sample Output

30
Source

IOI 95

本题是一道简单的DP问题。就是所谓的数塔问题。

下面附上我的AC代码。

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int dp[1005][1005];int main(){    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<n;i++)        {            for(int j=0;j<=i;j++)            {                scanf("%d",&dp[i][j]);            }        }        for(int i=n-2;i>=0;i--)        {            for(int j=0;j<=i;j++)            {                dp[i][j]=dp[i][j]+max(dp[i+1][j],dp[i+1][j+1]);            }        }        printf("%d\n",dp[0][0]);    }    return 0;}