poj 3126 Prime Path （bfs)

一学期前看题解敲的：

#include<iostream>
#include<queue>
#include<cstring>
#include<cstdlib>
#include<cmath>

using namespace std;
int n;
int prime[10010] = {0};
int book[10010];
int step[10010];

int bfs(int sou, int des) {
	queue<int> q;
	memset(book, 0, sizeof(book));
	memset(step, 0, sizeof(step));
	q.push(sou);
	book[sou] = 1;
	while(!q.empty()) {
		int x = q.front();
		q.pop();
		int t[4];
		t[0] = x / 1000;
		t[1] = x % 1000 / 100;
		t[2] = x % 100 / 10;
		t[3] = x % 10;
		int i, j, tmp;
		for(i = 0; i < 4; i++) {
			tmp = t[i];
			for(j = 0; j < 10; j++) {
				if(tmp != j) {
					t[i] = j;
					int y = t[3] + t[2] * 10 + t[1] * 100 + t[0] * 1000;
					if(!book[y] && prime[y]) {
						q.push(y);
						step[y] = step[x] + 1;
						book[y] = 1;
					}
					if(y == des) return step[y];
				}
			}
			t[i] = tmp;
		}
		if(x == des) return step[x];
	}
	return -1;
}

int main() {
	prime[0] = prime[1] = 1;
	int i, j;
	for(i = 0; i < 10010; i++) {
		prime[i] = 1;
	}
	for(i = 2; i <= sqrt(10010.0); i++) {
		if(prime[i]) {
			for(j = i + i; j < 10010; j += i) {
				prime[j] = 0;
			}
		}
	}	
	for(i = 0; i < 1000; i++) {
		prime[i] = 0;
	}
	cin >> n;
	while(n--) {
		int a, b;
		cin >> a >> b;
		int t = bfs(a, b);
		if(t == -1) cout << "Impossible" << endl;
		else cout << t << endl;
	}	
	return 0;
}

一学期后自己写的：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;
int isprime[10010];
int vis[10010];
void init() {
	int i, j;
	for(i = 0; i < 10000; i++) {
		isprime[i] = 1;
	}
	isprime[0] = isprime[1] = 0;
	for(i = 2; i <= sqrt(10000); i++) {
		if(isprime[i] == 1) {
			for(j = i + i; j < 10000; j += i) {
				isprime[j] = 0;
			}
		}
	}
	for(i = 0; i <= 999; i++) {
		isprime[i] = 0;
	}
}
int a, b;
struct node {
	int num, step;
};
int bfs() {
	queue<node> q;
	node now, next;
	now.num = a, now.step = 0;
	q.push(now);
	vis[now.num] = 1;
	while(!q.empty()) {
		now = q.front();
		q.pop();
		int i, j;
		if(now.num == b) {
			return now.step;
		}
		for(i = 0; i < 4; i++) {
			for(j = 0; j <= 9; j++) {
				if(now.num % (int)pow(10, i + 1) / (int)pow(10, i) == j) continue;
				int t1 = now.num / (int)pow(10, i + 1);
				int t2 = now.num % (int)pow(10, i);
				next.num = t1 * (int)pow(10, i + 1) + j * (int)pow(10, i) + t2;
				next.step = now.step + 1;
				if(vis[next.num] == 0 && isprime[next.num] == 1) {
					vis[next.num] = 1;
					q.push(next);
				}
			}
		}
	}
	return -1;
}
int main() {
	init();
    int n;
    scanf("%d", &n);
    while(n--) {
    	memset(vis, 0, sizeof(vis));
	    scanf("%d %d", &a, &b);
	    printf("%d\n", bfs());
	}
    return 0;
}

说明还是有提高的。。。多复习。