Unidirectional TSP（dp）

Description

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 columni 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 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.

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

题目大意：

给你一个n * m的矩阵，每个矩阵的元素都有相应的值，找出一条路径，从一列的任意一点出发走到底m列的任意一点出去，可以走的三个方向如图所示，当在第一行时，往右上角走的话可以最直走到第n行(在第n行往右下走的时候同理)。走到最后求出值最小的一个路径每次所在行，要求输出字典序最小的一条路。

解题思路：

运用数字三角形的思想可以一直从前往后每个点都加上前一列能走到该点的最小值，然后一直到最后一行。这种方法到求字典序最小的路径的时候异常麻烦，自己没写出来，所以换一种思想，从最后一列一直加到第一列，每次存取字典序最小的方案，然后求出解。

代码如下：

#include<iostream>
#include<cstdio>
#include<map>
#include<math.h>
#include<cstring>
#include<algorithm>
using namespace std;

int main()
{
    int i, j,vis[20][110], m, n;
    int dp[20][110], Map[20][110];
    while(scanf("%d%d",&m,&n) != EOF)
    {
        memset(Map,0,sizeof(Map));
        memset(dp,0,sizeof(dp));
        for(i = 1; i <= m; i++)
        {
            for(j = 1; j <= n; j++)
            {
                scanf("%d",&Map[i][j]);
            }
        }

        for(i = n; i >= 1; i--)
        {
            for(j = 1 ; j <= m; j++)
            {
                dp[j][i] = Map[j][i] +  dp[j][i + 1];//先加上往前走的
                vis[j][i] = j;    // vis数组保存每次的路径
                if(j > 1 && (dp[j][i] >= Map[j][i] + dp[j - 1][i + 1])) //如果王右上走小于等于当前值都要更新，因为字典序较小
                {
                       dp[j][i] = Map[j][i] + dp[j - 1][i + 1];
                       vis[j][i] = j - 1;
                }
                if(j == m && (dp[j][i] >= Map[j][i] + dp[1][i +1]))//同理字典序最小要更新。
                {
                       dp[j][i] = Map[j][i] + dp[1][i + 1];
                       vis[j][i] = 1;
                }
                if(j < m && (dp[j][i] > Map[j][i] + dp[j + 1][i + 1]))//只能先判断这一个，因为如果先判断下面的j == 1的话字典序有一种情况不是最优
                {
                       dp[j][i] = Map[j][i] + dp[j + 1][i +1];
                       vis[j][i] = j + 1;
                }
                if(j == 1 && (dp[j][i] > Map[j][i] +dp[m][i + 1]))
                {
                       dp[j][i] = Map[j][i] + dp[m][i + 1];
                       vis[j][i] = m;
                }

            }
        }
        int besti;
        int minx;
        besti = 1;//找到最优的方案
        for(i = 2; i <= m; i++)
        {
            if(dp[i][1] < dp[besti][1])
            {
                besti = i;
            }
        }
        printf("%d",besti);
        minx = dp[besti][1];
        //依次往下找就出来了
        for(i = 1; i < n; i++)
        {
            printf(" %d",vis[besti][i]);
            besti = vis[besti][i];
        }
        printf("\n");
        printf("%d\n",minx);
    }
    return 0;
}