洛谷——P1216 [USACO1.5][IOI1994]数字三角形 Number Triangles

动态规划:
原题链接
这道题的动态转移方程是a[i][j]+=max(a[i-1][j],a[i-1][j-1])
当然,要注意边界问题

#include<iostream>
using namespace std;
int max(int a, int b) {
    return a > b ? a : b;
}
int main()
{
    int n;
    cin >> n;
    int a[1001][1001] = { 0 };
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++) {
            cin >> a[i][j];
            if (i == 1)
                a[i][j] += a[i - 1][j];
            else if (j == i)
                a[i][j] += a[i - 1][j - 1];
            else
                a[i][j] += max(a[i - 1][j], a[i - 1][j - 1]);
        }
    }
    int ans = 0;
    for (int i = 0; i <= n; i++) {
        ans = max(ans, a[n][i]);
    }
    cout << ans;
    return 0;
}

优化一下

#include<iostream>
#include<vector>
#include<climits>
using namespace std;
int max(int a, int b) {
    return a > b ? a : b;
}
int main()
{
    int n;
    cin >> n;
    int a[1001] = { 0 };
    for (int i = n; i >= 1; i--) {
        for (int j = i; j <= n; j++) {
            int t;
            cin>>t;
            a[j]=t+max(a[j+1],a[j]);
        }
    }
    int ans = 0;
    for (int i = 1; i <= n; i++) {
        ans = max(ans, a[i]);
    }
    cout << ans;
    return 0;
}
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