南阳18--The Triangle

The Triangle

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

 

描述

7
3 8
8 1 0
2 7 4 4
4 5 2 6 5
(Figure 1)
Figure 1 shows a number triangle. Write a program that calculates the highest sum of numbers 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.

 

输入

Your program is to read from standard input. The first line contains one integer N: the number of rows in the triangle. The following N lines describe the data of the triangle. The number of rows in the triangle is > 1 but <= 100. The numbers in the triangle, all integers, are between 0 and 99.

输出

Your program is to write to standard output. The highest sum is written as an integer.

样例输入

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

样例输出

30

//这个是经典的简单动态规划，典型的多段图。思路就是建立一个数组，由下向上动态规划，保存页子节点到当前节点的最大值： 

for(int i=num-2;i>=0;i--) //核心 简单的动規； 
{
    for(int j=0;j<=i;j++)
　　{
        dp[i][j]+=max(dp[i+1][j],dp[i+1][j+1]);
     }
}

#include<stdio.h>
int main()
{
    int n,i,j,num[110][110];
    scanf("%d",&n);
    for(i=0;i<n;i++)
    for(j=0;j<=i;j++)
    scanf("%d",&num[i][j]);
    for(i=n-2;i>=0;i--)
    for(j=0;j<=i;j++)
    {
        num[i][j]+=num[i+1][j]>num[i+1][j+1]?num[i+1][j]:num[i+1][j+1]; 
    }
    printf("%d
",num[0][0]);
    return 0;
}

重刷：

#include <cstdio>
#include <cstring>
#define max(a, b) (a>b?a:b)
const int N = 101;
int dp[N][N], num[N][N];
int main(){
    int n;
    while(scanf("%d", &n) != EOF){
        memset(dp, 0, sizeof(dp));
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= i; j++)
                scanf("%d", &num[i][j]);
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= i; j++)
                dp[i][j]=max(dp[i-1][j], dp[i-1][j-1])+num[i][j];
        int max = -1;
        for(int i = 1; i <= n; i++)
            if(dp[n][i] > max)
                max = dp[n][i];
        printf("%d
", max); 
    }
    return 0;
}