数字三角形模型

P1004 [NOIP2000 提高组] 方格取数

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 10;

int n;
int w[N][N];
int f[N][N][N];//f[i1, j1, i2, j2]
               //当i1 + j1 == i2 + j2时两点才可能重合
               //所以用k表示i和j的和优化掉一维
int main()
{
    cin.tie(0);
    cin >> n;
    int a, b, c;
    while (cin >> a >> b >> c, a || b || c) w[a][b] = c;

    for (int k = 2; k <= n + n; k++)//k = i1 + j1 = i2 + j2 从（1，1）点开始
        for (int i1 = 1; i1 <= n; i1++)
            for (int i2 = 1; i2 <= n; i2++)
            {
                int j1 = k - i1, j2 = k - i2;
                if(j1 >= 1 && j1 <= n && j2 >= 1 && j2 <= n)
                {
                    int t = w[i1][j1];
                    if (i1 != i2)//当i1 == i2时,j1 = k - i1, j2 = k - i2 = k - i1,此时两点重合
                        t += w[i2][j2];

                    f[k][i1][i2] = max(f[k][i1][i2], f[k - 1][i1 - 1][i2 - 1] + t);//都从上来
                    f[k][i1][i2] = max(f[k][i1][i2], f[k - 1][i1 - 1][i2] + t);//第一条从上来,第二条从左来
                    f[k][i1][i2] = max(f[k][i1][i2], f[k - 1][i1][i2 - 1] + t);//第二条从上来,第一条从左来
                    f[k][i1][i2] = max(f[k][i1][i2], f[k - 1][i1][i2] + t);//都从左来
                }
            }
    cout << f[n + n][n][n] << endl;

    return 0;
}