UVA - 116 Unidirectional TSP

Input
The input consists of a sequence of matrix specifications. Each matrix specification consists of the row
and column dimensions in that order on a line followed by m · n integers where m is the row dimension
and n is the column dimension. The integers appear in the input in row major order, i.e., the first n
integers constitute the first row of the matrix, the second n integers constitute the second row and so
on. The integers on a line will be separated from other integers by one or more spaces. Note: integers
are not restricted to being positive.
There will be one or more matrix specifications in an input file. Input is terminated by end-of-file.
For each specification the number of rows will be between 1 and 10 inclusive; the number of columns
will be between 1 and 100 inclusive. No path’s weight will exceed integer values representable using 30
bits.

Output
Two lines should be output for each matrix specification in the input file, the first line represents a
minimal-weight path, and the second line is the cost of a minimal path. The path consists of a sequence
of n integers (separated by one or more spaces) representing the rows that constitute the minimal path.
If there is more than one path of minimal weight the path that is lexicographically smallest should be
output.
Note: Lexicographically means the natural order on sequences induced by the order on their elements.
Sample Input
5 6
3 4 1 2 8 6
6 1 8 2 7 4
5 9 3 9 9 5
8 4 1 3 2 6
3 7 2 8 6 4
5 6
3 4 1 2 8 6
6 1 8 2 7 4
5 9 3 9 9 5
8 4 1 3 2 6
3 7 2 1 2 3
2 2
9 10 9 10

Sample Output
1 2 3 4 4 5
16
1 2 1 5 4 5
11
1 1
19

 

首先思考用暴力做，发现大问题和小问题有交叉的关系，所以思考用递归做，进而得出状态转移方程

#include <iostream>
#include <algorithm>

using namespace std;
#define N 101
#define INF 0x7fffffff //最高位不要取f，会变成1哒


int main() {
    int A[N][N], d[N][N], p[N][N], n, m, ans, first;

    while (cin >> m >> n) {
        ans = INF;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                cin >> A[i][j];
            }
        }
        for (int j = n - 1; j >= 0; --j) {
            for (int i = 0; i < m; ++i) {
                if (j == n - 1)
                    d[i][j] = A[i][j];
                else {
                    int rows[] = {i - 1, i, i + 1};
                    if (i == 0) rows[0] = m - 1;
                    if (i == m - 1) rows[2] = 0;
                    sort(rows, rows + 3); //利用排序来取索引和最小值
                    int v = INF;
                    for (int k = 0; k < 3; ++k) {
                        if (v > d[rows[k]][j + 1]) {
                            v = d[rows[k]][j + 1];
                            p[i][j] = rows[k]; //记录下一步的路径
                        }
                    }
                    d[i][j] = v + A[i][j];
                }
                if (j == 0 and ans > d[i][j]) {//由上往下扫，形成字典序
                    ans = d[i][j], first = i;
                }
            }
        }
        cout << first + 1;
        for (int j = 0, r = first; j < n - 1; ++j) {//路径的输出
            r = p[r][j];
            printf(" %d", r + 1);
        }
        printf("\n%d\n", ans);
    }
}