UVa - 116 - Unidirectional TSP

Background

Problems that require minimum paths through some domain appear in many different areas of computer science. For example, one of the constraints in VLSI routing problems is minimizing wire length. The Traveling Salesperson
Problem (TSP) -- finding whether all the cities in a salesperson's route can be visited exactly once with a specified limit on travel time -- is one of the canonical examples of an NP-complete problem; solutions appear to require an inordinate amount of time
to generate, but are simple to check.

This problem deals with finding a minimal path through a grid of points while traveling only from left to right.

The Problem

Given an  matrix of integers, you are to write a program that computes a path of
minimal weight. A path starts anywhere in column 1 (the first column) and consists of a sequence of steps terminating in column n (the last column). A step consists of traveling from column i to column i+1 in an adjacent (horizontal
or diagonal) row. The first and last rows (rows 1 and m) of a matrix are considered adjacent, i.e., the matrix ``wraps'' so that it represents a horizontal cylinder. Legal steps are illustrated below.

The weight of a path is the sum of the integers in each of the n cells of the matrix that are visited.

For example, two slightly different  matrices are shown below (the only difference
is the numbers in the bottom row).

The minimal path is illustrated for each matrix. Note that the path for the matrix on the right takes advantage of the adjacency property of the first and last rows.

The 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  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.

The 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 nintegers
(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.

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

动态规划问题，倒着找一遍即可，不过这个要输出结果，所以需要记录下结果。不过AC了之后等级不高，速度稍慢。

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cctype>
#include <cstring>
#include <string>
#include <sstream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>
#include <stack>
#include <queue>
#include <bitset>
#include <cassert>
#include <cmath>
#include <functional>

using namespace std;

const int maxm = 12;
const int maxn = 105;
const int inf = 0x3f3f3f3f;

int matrix[maxm][maxn], nextLine[maxm][maxn];
int d[maxm][maxn]; // d[i][j]表示从格子(i, j)出发到最后一列的最小开销
int m, n;

void init()
{
	cin >> n;
	for (int i = 0; i < m; i++) {
		for (int j = 0; j < n; j++) {
			cin >> matrix[i][j];
		}
	}
}

// 动态规划。。。
void solve()
{
	int ans = inf, first = 0;
	for (int j = n - 1; j >= 0; j--) { // 逆推
		for (int i = 0; i < m; i++) {
			if (j == n - 1) { // 边界
				d[i][j] = matrix[i][j];
			}
			else {
				int row[3] = { i, i - 1, i + 1 };
				if (i == 0) {
					row[1] = m - 1; // 第0行上面是第m-1行
				}
				if (i == m - 1) {
					row[2] = 0; // 第m-1行下面是第0行
				}
				sort(row, row + 3); // 重排序是为了得到字典序最小的
				d[i][j] = inf;
				for (int k = 0; k < 3; k++) { // 向左走，判断那个比较小
					int v = d[row[k]][j + 1] + matrix[i][j];
					if (v < d[i][j]) {
						d[i][j] = v;
						nextLine[i][j] = row[k];
					}
				}
			}
			if (j == 0 && d[i][j] < ans) {
				ans = d[i][j];
				first = i;
			}
		}
	}
	// 输出
	cout << first + 1;
	for (int i = nextLine[first][0], j = 1; j < n; i = nextLine[i][j], j++) {
		cout << ' ' << i + 1;
	}
	cout << endl << ans << endl;
}

int main()
{
	ios::sync_with_stdio(false);
	while (cin >> m) {
		init();
		solve();
	}

	return 0;
}