HDOJ1016.Prime Ring Problem

试题请参见: http://acm.hdu.edu.cn/showproblem.php?pid=1016

题目概述

A ring is compose of n circles as shown in diagram. Put natural number 1, 2, …, n into each circle separately, and the sum of numbers in two adjacent circles should be a prime.

Note: the number of first circle should always be 1.

解题思路

回溯. 生成以1作为第1个元素的全排列.

遇到的问题

TLE. 于是想到了一个剪枝的方法, 如果当前的序列(如

1 3

)已经不能构成质数环(因为1 + 3 = 4不是质数), 则不要继续搜索.

PE. 输出结果的时候, 若是序列中的最后一个数字, 不需要在数字之后加空格.

源代码

#include <iostream>
#include <cmath>

bool isPrime(int x) {
for ( int i = 2; i <= std::sqrt(x); ++ i ) {
if ( x % i == 0 ) {
return false;
}
}
return true;
}

bool isPrimeLine(int* ring, int n) {
for ( int i = 1; i < n; ++ i ) {
if ( !isPrime(ring[i] + ring[i - 1]) ) {
return false;
}
}
return true;
}

bool isPrimeRing(int* ring, int n) {
for (int i = 0; i < n; ++i) {
if (!isPrime(ring[(i + n) % n] + ring[(i + n + 1) % n])) {
return false;
}
}
return true;
}

void getPrimeRing(int* ring, bool* isUsed, int selectedNumbers, int totalNumbers) {
if ( selectedNumbers == totalNumbers ) {
if ( isPrimeRing(ring, totalNumbers) ) {
for ( int i = 0; i < totalNumbers; ++ i ) {
std::cout << ring[i] << ( i == totalNumbers - 1 ? "\n" : " " );
}
}
}

for ( int i = 1; i < totalNumbers; ++ i ) {
if ( !isUsed[i] ) {
isUsed[i] = true;
ring[selectedNumbers] = i + 1;

if ( isPrimeLine(ring, selectedNumbers + 1) ) {
getPrimeRing(ring, isUsed, selectedNumbers + 1, totalNumbers);
}
isUsed[i] = false;
}
}
}

int main() {
const int MAX_SIZE = 20;

// Put number 1 as the first element
bool isUsed[MAX_SIZE] = { 1, 0 };
int ring[MAX_SIZE] = { 1, 0 };
int n = 0, currentCase = 0;

while ( std::cin >> n ) {
std::cout << "Case " << ++ currentCase << ":" << std::endl;
getPrimeRing(ring, isUsed, 1, n);
std::cout << std::endl;
}

return 0;
}