UVA 116 Unidirectional TSP

//  [解题方法]
//  记忆化搜索（递归，子问题的结果用备忘录存起来，避免重复求解）
//  二维nxt数组按照题意记录路径
//  dp[x][y](即dfs(x,y))表示从(x,y)走到最右边需要的最小花费

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
using namespace std;
#define LL long long
#define inf 1e16

LL dp[11][111],w[11][111], mins, res[105];
int r, c;

struct point{
	int x, y;
}nxt[11][111];

LL dfs (int x, int y)
{
	if (y >= c) return (dp[x][y]=w[x][y]);
	if (dp[x][y] < inf)
		return dp[x][y];
	int i;
	LL tp = inf;
	for (i = x-1; i <= x+1; i++)
	{
		int tr = i;
		if (tr == 0) tr = r;
		else if (tr > r) tr = 1;
		LL tmp = dfs (tr, y+1) + w[x][y];
		//更新路径(花费更小+字典序)
		if (tmp < tp || (tmp == tp && tr < nxt[x][y].x)) {
			nxt[x][y].x = tr;
			nxt[x][y].y = y+1;
			tp = tmp;
		}
	}
	return (dp[x][y]=tp);
}

int main()
{
	int i, j;
	while (cin >> r >> c)
	{
		for (i = 1; i <= r; i++)
			for (j = 1; j <= c; j++)
				cin >> w[i][j], dp[i][j] = inf;
		mins = inf;
		int v;
		memset (nxt, -1, sizeof(nxt));
		for (i = 1; i <= r; i++)
		{
			LL tp = dfs (i, 1);
			if (tp < mins) {
				mins = tp;
				v = i;
			}
		}
		int tc = 1, k = 0;
		while (v > 0)
		{
			res[k++] = v;
			int tv = v;
			v = nxt[tv][tc].x;
			tc = nxt[tv][tc].y;
		}
		cout << res[0];
		for (i = 1; i < k; i++) cout << ' ' << res[i];
		cout << endl << mins << endl;
	}
    return 0;
}