UVa 116 Unidirectional TSP（简单旅行商DP）

题意：

求代价最小的一条路径。

思路：

由于要求路径，并且要求输出字典顺序最小的序列。所以逆向求DP，模仿dfs，将降低解题难度。

#include <cstdio>
#include <cstdlib>
#include <cstring>

#define min(a,b) (((a) < (b)) ? (a) : (b))

int map[15][105];
int dp[15][105], path[15][105];
int n, m;

int main()
{
    while (scanf("%d %d", &n, &m) != EOF)
    {
        for (int i = 0; i < n; ++i)
            for (int j = 0; j < m; ++j)
                scanf("%d", &map[i][j]);

        memset(dp, 0, sizeof(dp));
        memset(path, 0,sizeof(path));

        for (int j = m - 1; j >= 0; --j)
        {
            for (int i = 0; i < n; ++i)
            {
                int a = dp[(i-1+n)%n][j+1];
                int b = dp[i][j+1];
                int c = dp[(i+1)%n][j+1];
                int mm;
                if (a > b)
                    mm = min(b, c);
                else
                    mm = min(a, c);

                dp[i][j] = map[i][j] + mm;
                
                path[i][j] = 1e9;

                if (mm == a)
                    path[i][j] = (i - 1 + n) % n;

                if (mm == b)
                    path[i][j] = min(path[i][j], i);

                if (mm == c)
                    path[i][j] = min(path[i][j], (i + 1) % n);
            }
        }
        int ans = 1e9;
        int id;
        for (int i = 0; i < n; ++i)
            if (ans > dp[i][0])
                ans = dp[i][0], id = i;

        printf("%d", id + 1);
        id = path[id][0];

        for (int j = 1; j < m; ++j)
        {
            printf(" %d", id + 1);
            id = path[id][j];
        }

        printf("\n%d\n", ans);
    }
    return 0;
}