NYOJ-The Triangle【数塔问题】

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

573 88 1 0 2 7 4 44 5 2 6 5

样例输出

30

思路:

就是数塔问题，每一个可以向左下或者右下走，问走到低时路过数字最大和是多少，所以我们可以从倒数第二层开始向上找，所以状态转移方程为dp[i][j]+=max(dp[i+1][j],dp[i+1][j+1]);，最后输出第一个位置的值就是最大值。

代码：

#include <bits/stdc++.h>using namespace std;int main(){    int dp[105][105];    int w;cin>>w;    memset(dp,0,sizeof(dp));    for(int i=1;i<=w;i++)        for(int j=1;j<=i;j++)    {        scanf("%d",&dp[i][j]);    }    for(int i=w-1;i>0;i--)        for(int j=1;j<=i;j++)    {        dp[i][j]+=max(dp[i+1][j],dp[i+1][j+1]);    }    cout<<dp[1][1]<<endl;    return 0;}