# 程序控

IPPP (Institute of Penniless Peasent-Programmer) Fellow

:: :: :: :: :: :: :: ::
 77 随笔 :: 0 文章 :: 457 评论 :: 0 引用

Time limit: 3.000 seconds

## Background背景

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问题

Given an m × n 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 5 × 6 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输入

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 m × n 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输出

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.

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

1 2 3 4 4 5
16
1 2 1 5 4 5
11
1 1
19

## Solution解答

#include <algorithm>
#include <iostream>
using namespace std;
//主函数
int main(void) {
//aMat用于存储距阵，aPrev存储上一列(前一步)的行号
//为方便运算，这里采用列在前行在后的维度排列
int aMat[100][10], aPrev[100][10];
//循环处理所有矩阵
for (int nRows, nCols; cin >> nRows >> nCols; ) {
//循环读入矩阵
for (int i = 0; i < nRows; ++i) {
for (int j = 0; j < nCols; cin >> aMat[j++][i]);
}
//从后向前循环遍例每一例
for (int j = nCols - 2; j >= 0; --j) {
for (int i = 0; i < nRows; ++i) {
//为加快速度，减少代码，为数组元素设置临时指针变量
int *p = &aMat[j + 1][0];
//设最小值所在行号为上面一行(右上方一格)
int nMin = (i - 1 + nRows) % nRows, d = (i + 1) % nRows;
//与右方一格比较，取最小值。若相等按字母顺序排列
if (p[i] < p[nMin] || (p[i] == p[nMin] && i < nMin)) {
nMin = i;
}
//与右下方一格比较，取最小值。若相等按字母顺序排列
if (p[d] < p[nMin] || (p[d] == p[nMin] && d < nMin)) {
nMin = d;
}
//更新当前一列的值为从最末一列到当前列的最小值
aMat[j][i] += p[nMin];
//记录上一步的行号
aPrev[j][i] = nMin;
}
}
//检查开头一列的最小值，即最小路径总权重
int nMin = min_element(&aMat[0][0], &aMat[0][nRows]) - &aMat[0][0];
int nWeight = aMat[0][nMin];
//避免在行尾输出多余空格，先输出第1个值
cout << nMin + 1;
//循环输出路径(各列行号)
for (int j = 1; j < nCols; ++j) {
//获取下一步的行号并输出
nMin = aPrev[j - 1][nMin];
cout << ' ' << nMin + 1;
}
//输出路径总权重
cout << '\n' << nWeight << endl;
}
return 0;
}
posted on 2010-08-13 02:37  Devymex  阅读(3156)  评论(1编辑  收藏