Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1388 Accepted Submission(s): 866
Problem Description
The ministers of the cabinet were quite upset by the message from the Chief of Security stating that they would all have to change the four-digit room numbers on their offices.
— It is a matter of security to change such things every now and then, to keep the enemy in the dark.
— But look, I have chosen my number 1033 for good reasons. I am the Prime minister, you know!
—I know, so therefore your new number 8179 is also a prime. You will just have to paste four new digits over the four old ones on your office door.
— No, it’s not that simple. Suppose that I change the first digit to an 8, then the number will read 8033 which is not a prime!
— I see, being the prime minister you cannot stand having a non-prime number on your door even for a few seconds.
— Correct! So I must invent a scheme for going from 1033 to 8179 by a path of prime numbers where only one digit is changed from one prime to the next prime.
Now, the minister of finance, who had been eavesdropping, intervened.
— No unnecessary expenditure, please! I happen to know that the price of a digit is one pound.
— Hmm, in that case I need a computer program to minimize the cost. You don’t know some very cheap software gurus, do you?
—In fact, I do. You see, there is this programming contest going on. . .
Help the prime minister to find the cheapest prime path between any two given four-digit primes! The first digit must be nonzero, of course. Here is a solution in the case above.
1033
1733
3733
3739
3779
8779
8179
The cost of this solution is 6 pounds. Note that the digit 1 which got pasted over in step 2 can not be reused in the last step – a new 1 must be purchased.
Input
One line with a positive number: the number of test cases (at most 100). Then for each test case, one line with two numbers separated by a blank. Both numbers are four-digit primes (without leading zeros).
Output
One line for each case, either with a number stating the minimal cost or containing the word Impossible.
Sample Input
3
1033 8179
1373 8017
1033 1033
Sample Output
6
7
0
Source
NWERC2006
问题链接:HDU1973 POJ3126 UVA12101 LA3639 Prime Path
问题简述:(略)
问题分析:
求素数最短路径问题,每个结点是四位数的素数,结点间一位不同则有边相连。最短路径问题通常用BFS来解决。
预先用筛选法求得所有四位数的素数备用,可以加速。出现过的四位数素数做个标记,不用重复搜索。
程序说明:
这个解题程序代码可以改进,可以用2重循环来替代,可以简化代码。
这个程序在UVALive中提交出现WA。
参考链接:(略)
题记:(略)
AC的C++语言程序如下:
/* HDU1973 POJ3126 UVA12101 LA3639 Prime Path */
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <queue>
using namespace std;
const int N = 10000;
const int SQRTN = sqrt((double) N);
bool isprime[N + 1];
// Eratosthenes筛选法
void esieve(void)
{
memset(isprime, true, sizeof(isprime));
isprime[0] = isprime[1] = false;
for(int i = 2; i <= SQRTN; i++) {
if(isprime[i]) {
for(int j = i * i; j <= N; j += i) //筛选
isprime[j] = false;
}
}
}
int vis[N];
int cnt, sp, ep;
void bfs(int p)
{
memset(vis, 0, sizeof(vis));
queue<pair<int, int> > q;
q.push(make_pair(0, p));
while(!q.empty()) {
pair<int, int> t = q.front(), n;
q.pop();
if(t.second == ep) {
cnt = t.first;
return;
}
// 个位,只能是奇数
int d1 = t.second % 10;
for(int i = 1; i < 10; i += 2) {
if(i != d1) {
n = t;
n.second -= d1;
n.second += i;
if(isprime[n.second] && vis[n.second] == 0) {
vis[n.second] = 1;
n.first++;
q.push(n);
}
}
}
// 十位
d1 = t.second / 10 % 10;
for(int i = 0; i < 10; i++) {
if(i != d1) {
n = t;
n.second -= d1 * 10;
n.second += i * 10;
if(isprime[n.second] && vis[n.second] == 0) {
vis[n.second] = 1;
n.first++;
q.push(n);
}
}
}
// 百位
d1 = t.second / 100 % 10;
for(int i = 0; i < 10; i++) {
if(i != d1) {
n = t;
n.second -= d1 * 100;
n.second += i * 100;
if(isprime[n.second] && vis[n.second] == 0) {
vis[n.second] = 1;
n.first++;
q.push(n);
}
}
}
// 千位,不能是0
d1 = t.second / 1000;
for(int i = 1; i < 10; i++) {
if(i != d1) {
n = t;
n.second -= d1 * 1000;
n.second += i * 1000;
if(isprime[n.second] && vis[n.second] == 0) {
vis[n.second] = 1;
n.first++;
q.push(n);
}
}
}
}
}
int main()
{
esieve();
int t;
scanf("%d\n", &t);
while(t--) {
scanf("%d%d", &sp, &ep);
cnt = -1;
bfs(sp);
if(cnt == -1) printf("Impossible\n");
else printf("%d\n", cnt);
}
return 0;
}