# 程序控

IPPP (Institute of Penniless Peasent-Programmer) Fellow

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Time limit: 3.000 seconds

## Problem问题

Given a graph (V,E) where V is a set of nodes and E is a set of arcs in VxV, and an ordering on the elements in V, then the bandwidth of a node v is defined as the maximum distance in the ordering between v and any node to which it is connected in the graph. The bandwidth of the ordering is then defined as the maximum of the individual bandwidths. For example, consider the following graph:

This can be ordered in many ways, two of which are illustrated below:

For these orderings, the bandwidths of the nodes (in order) are 6, 6, 1, 4, 1, 1, 6, 6 giving an ordering bandwidth of 6, and 5, 3, 1, 4, 3, 5, 1, 4 giving an ordering bandwidth of 5.

Write a program that will find the ordering of a graph that minimises the bandwidth.

## Input输入

Input will consist of a series of graphs. Each graph will appear on a line by itself. The entire file will be terminated by a line consisting of a single#. For each graph, the input will consist of a series of records separated by ;'. Each record will consist of a node name (a single upper case character in the the range A' to Z'),followed by a :' and at least one of its neighbours. The graph will contain no more than 8 nodes.

## Output输出

Output will consist of one line for each graph, listing the ordering of the nodes followed by an arrow (->) and the bandwidth for that ordering. All items must be separated from their neighbours by exactly one space. If more than one ordering produces the same bandwidth, then choose the smallest in lexicographic ordering, that is the one that would appear first in an alphabetic listing.

## Sample input示例输入

A:FB;B:GC;D:GC;F:AGH;E:HD
#

## Solution解答

#include "stdafx.h"
#include <algorithm>
#include <iostream>
#include <vector>
#include <string>

typedef std::pair<char, char> NODEPAIR;
typedef std::vector<char>::iterator NODE_ITER;
typedef std::string::iterator STR_ITER;
typedef std::vector<NODEPAIR>::iterator GRAPH_ITER;

int main(void)
{
char idxTbl[32]; // 从顶点编号到其在某种排序中的位置的对应表
std::string strLine; // 存储一行输入的字符串
for (; std::getline(std::cin, strLine) && strLine[0] != '#'; ) {
std::vector<NODEPAIR> graph; // 数组中每个元素为一条边，用一对顶点的编号表示
std::vector<char> nodes; // 记录一种顶点的排序
// 为方便判断一行输入的结束，在行层添加分号
strLine.push_back(';');
// 循环处理当前输入行的每一个字符
for (STR_ITER i = strLine.begin(); i != strLine.end(); ++i) {
// 用所有边的顶点对来表示图，nFrom是冒号前的顶点，nTo是冒号后的顶点
char nFrom = *i - 'A'; // 每个顶点的编号为其字母的ASCII码-'A'
// 有些顶点只出现在冒号前，有些只出现在冒号后，因此nFrom和nTo都需添加
nodes.push_back(nFrom);
// 遍历冒号后面直到分号的顶点，这些顶点用nTo表示
for (i += 2; *i != ';'; ++i) {
char nTo = *i - 'A';  // 每个顶点的编号为其字母的ASCII码-'A'
nodes.push_back(nTo);
// 保证添加到图中的顶点对（边）中编号较小的顶点在前，避免无向图的重复边
if (nFrom > nTo)
graph.push_back(NODEPAIR(nFrom, nTo));
else if (nFrom < nTo)
graph.push_back(NODEPAIR(nTo, nFrom));
}
}
// 对图的所有顶点对（边）排序去重
std::sort(graph.begin(), graph.end());
graph.erase(std::unique(graph.begin(), graph.end()), graph.end());
// 对顶点数组排序去重，作为第一种排序（升序最小）
std::sort(nodes.begin(), nodes.end());
nodes.erase(std::unique(nodes.begin(), nodes.end()), nodes.end());
std::vector<char> minOrder; // 记录具有最小带宽的排序
char nMinBw = char(nodes.size()); // 记录具有最小的带宽，初始化为数组长度
for (bool bNext = true; bNext; ) { // 遍历所有排序
char nCnt = 0, nOrderBw = 0;
// 扫描一遍当前的排序，求出每一个顶点在当前排序中的位置
for (NODE_ITER i = nodes.begin(); i != nodes.end(); ++i) {
idxTbl[*i] = nCnt++;
}
// 遍历图中的所有顶点对，找出在当前排序中距离最远的顶点对作为当前排序的带宽
for (GRAPH_ITER i = graph.begin(); i != graph.end(); ++i) {
char nCur = char(std::abs(idxTbl[i->first] - idxTbl[i->second]));
if (nCur > nOrderBw)
nOrderBw = nCur;
}
// 如果当前排序的带宽小于已知最小带宽，则更新最小带宽和最小排序
if (nOrderBw < nMinBw) {
nMinBw = nOrderBw;
minOrder = nodes;
}
// 接字母序给出下一种排序
bNext = std::next_permutation(nodes.begin(), nodes.end());
}
// 按格式循环输出最小排序和它的带宽值
for (NODE_ITER i = minOrder.begin(); i != minOrder.end(); ++i) {
std::cout << char(*i + 'A') << ' ';
}
std::cout << "-> " << (int)nMinBw << std::endl;
}
return 0;
}


posted on 2013-08-17 15:26  Devymex  阅读(2078)  评论(2编辑  收藏