AcWing 898. 数字三角形
原题链接
简单
作者:
我要出去乱说
,
2021-01-27 20:36:41
,
所有人可见
,
阅读 273
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 510, INF = 1e9;
int n;
int a[N][N]; //a存三角形
int f[N][N]; //f存状态
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i ++ )
for (int j = 1; j <= i; j ++ )
scanf("%d", &a[i][j]);
for (int i = 0; i <= n; i ++ )
for (int j = 0; j <= i + 1; j ++ ) //每行要多初始化左右各一个(考虑如右斜边的右上方元素)
f[i][j] = -INF;
f[1][1] = a[1][1];
for (int i = 2; i <= n; i ++ )
for (int j = 1; j <= i; j ++ )
//状态转移方程:当前点最大值=左上节点最大值加当前点或右上节点最大值加当前点
f[i][j] = max(f[i - 1][j - 1] + a[i][j], f[i - 1][j] + a[i][j]);
int res = -INF;
for (int i = 1; i <= n; i ++ ) res = max(res, f[n][i]); //最后一行扫尾
printf("%d\n", res);
return 0;
}