dp算法之平安果路径问题c++

前文：https://www.cnblogs.com/ljy1227476113/p/9563101.html

在此基础上更新了可以看到行走路径的代码。

代码：

#include <iostream>
using namespace std;
int ivec[10001][10001];
int dp[10001][10001];
int que[10001];
int main()
{
    int m, n;
    int i, j;
    int tail = 1;
    while (cin >> m >> n)
    {
        que[10001] = { 0 };
        for (i = 1; i <= m; i++)
        {
            for (j = 1; j <= n; j++)
            {
                cin >> ivec[i][j];
            }
        }
        que[0] = ivec[m][n];
        dp[1][1] = ivec[1][1];
        for (i = 2; i <= m; i++)
        {
            dp[i][1] = dp[i - 1][1]+ivec[i][1];
        }
        for (j = 2; j <= n; j++)
        {
            dp[1][j] = dp[1][j - 1]+ivec[1][j];
        }
        for (i = 2; i <= m; i++)
        {
            for (j = 2; j <= n; j++)
            {
                dp[i][j] = ((dp[i - 1][j] < dp[i][j - 1]) ? dp[i][j - 1] : dp[i - 1][j]) + ivec[i][j];
            }
        }
        i = m, j = n;
        while(tail <= m + n - 2)
        {
            if (dp[i - 1][j] >= dp[i][j - 1])
            {
                que[tail++] = ivec[i - 1][j];
                i--;
            }
            else
            {
                que[tail++] = ivec[i][j - 1];
                j--;
            }
        }
        for (i = 1; i <= m; i++)
        {
            for (j = 1; j <= n; j++)
                cout << dp[i][j] << " ";
            cout << endl;
        }
        for (i = m + n - 2; i >= 0; i--)
            cout << que[i] << " ";
        cout << endl;
        cout << dp[m][n] << endl;
    }
    return 0;
}

结果：

输入：

2 4

1 2 3 40

6 7 8 90

输出：

1 2 3 40 90

136