UVA116多段图的最短路

Unidirectional TSP

Time Limit: 3000MS

Memory Limit: Unknown

64bit IO Format: %lld & %llu

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 TravelingSalesperson 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 inordinateamount 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 anmatrix of integers, you are to write a programthat 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+1in 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 differentmatrices 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 byintegerswhere 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 secondrow 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 sequenceof 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

/*Sample Input5 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 10Sample Output1 2 3 4 4 5
16
1 2 1 5 4 5
11
1 1
19题意：输入一个整形矩阵，从第一列任意位置出发每次往右，右上，右下走一格，最终到达最后一列要求经过的整形之和最小，矩阵是环形，即第一行的上一行是最后一行，最后一行的下一行是第一行。输出路径上面每一列的行号，如果存在多解，输出字典序最小的。*/// UVa116 Unidirectional TSP// 环形矩阵// 算法：多段图的动态规划。因为要字典序最小，所以倒着来，设d[i][j]为从(i,j)到最后一列的最小开销，则d[i][j]=a[i][j]+max(d[i+1][j+1],d[i-1][j+1])#include<cstdio>#include<algorithm>using namespace std;const int maxn = 100 + 5;const int INF = 1000000000;int m, n, a[maxn][maxn], d[maxn][maxn], next[maxn][maxn];int main(){///注意输入语句while(scanf("%d%d", &m, &n) == 2 && m)  ///m表示行 n表示列{for(int i = 0; i < m; i++)for(int j = 0; j < n; j++)scanf("%d", &a[i][j]);int ans = INF, first = 0;///从最后一列往前递推for(int j = n-1; j >= 0; j--)///对应列{for(int i = 0; i < m; i++)///对应行{if(j == n-1) d[i][j] = a[i][j];///边界else{int rows[3] = {i, i-1, i+1};///对应下一步可以走的三个行///注意要判断第i行是否属于第0行或者第m-1行if(i == 0) rows[1] = m-1;   ///如果等于0的时候，则它的下一行是最后一行，即m-1if(i == m-1) rows[2] = 0;   ///此时它的下一行是第一行sort(rows, rows+3); ///从小到大，先取最小行，要满足字典序d[i][j] = INF;for(int k = 0; k < 3; k++){int v = d[rows[k]][j+1] + a[i][j];///用上一列所求的最小值去更新现在的v值if(v < d[i][j])///寻找值最小的一个{d[i][j] = v;next[i][j] = rows[k];///存取的是行}}}if(j == 0 && d[i][j] < ans)///到达第0列，而且现在所求的值比原来的值还要小，对原来的值进行更新{ans = d[i][j];first = i;///记录行}}}printf("%d", first+1);for(int i = next[first][0], j = 1; j < n; i = next[i][j], j ++)///j对应列printf(" %d", i+1);printf("\n%d\n", ans);}return 0;}