暑假编程训练_数塔取数

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数塔取数

这是一道入门的动态规划题, 没做过的可能很难想出, 但是做过的会觉得很简单.

这道题相比最基础的数塔取数多了一个求路径. 只要记录所继承的下一层数的下标. 最后从头开始走过这些下标就是路径了.

时间复杂度是O(n), n为数塔的数的个数.

复制代码

#include <iostream>
#include <cstring>
using namespace std;

const int maxn = 1005;
int n, sum[maxn][maxn];

struct T {
    int x, y;
} arr[maxn][maxn];

int main()
{
    memset(sum, 0, sizeof(sum));
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= i; ++j) {
            cin >> arr[i][j].x;
        }
    }
    for (int i = 1; i <= n; ++i) {
        arr[n][i].y = i;
        sum[n][i] = arr[n][i].x;
    }

for (int i = n - 1; i >= 1; --i) {
        for (int j = 1; j <= i; ++j) {
            if (sum[i + 1][j] > sum[i + 1][j + 1]) {
                sum[i][j] = sum[i + 1][j] + arr[i][j].x;
                arr[i][j].y = j;
            } else {
                sum[i][j] = sum[i + 1][j + 1] + arr[i][j].x;
                arr[i][j].y = j + 1;
            }
        }
    }

/*for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= i; ++j) {
            cout << sum[i][j] << ' ';
        }
        cout << endl;
    }*/

cout << sum[1][1] << 'n';

int k = 1, next = 1;
    while (k <= n) {
        cout << arr[k][next].x << ' ';
        next = arr[k++][next].y;
    }
    cout << endl;
}

/*
5

9
12 15
10 6 8
2 18 9 5
19 7 10 4 16
*/

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#include <iostream>
#include <cstring>
using namespace std;

const int maxn = 1005;
int n, sum[maxn][maxn];

struct T {
    int x, y;
} arr[maxn][maxn];

int main()
{
    memset(sum, 0, sizeof(sum));
    cin >> n;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= i; ++j) {
            cin >> arr[i][j].x;
        }
    }
    for (int i = 1; i <= n; ++i) {
        arr[n][i].y = i;
        sum[n][i] = arr[n][i].x;
    }

    for (int i = n - 1; i >= 1; --i) {
        for (int j = 1; j <= i; ++j) {
            if (sum[i + 1][j] > sum[i + 1][j + 1]) {
                sum[i][j] = sum[i + 1][j] + arr[i][j].x;
                arr[i][j].y = j;
            } else {
                sum[i][j] = sum[i + 1][j + 1] + arr[i][j].x;
                arr[i][j].y = j + 1;
            }
        }
    }

    /*for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= i; ++j) {
            cout << sum[i][j] << ' ';
        }
        cout << endl;
    }*/

    cout << sum[1][1] << 'n';

    int k = 1, next = 1;
    while (k <= n) {
        cout << arr[k][next].x << ' ';
        next = arr[k++][next].y;
    }
    cout << endl;
}

/*
5

9
12 15
10 6 8
2 18 9 5
19 7 10 4 16
*/