poj 3126 Prime Path

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简单广搜，练练代码把握能力，不能水题也要调半天了。。。。。

先将四位的素数筛选出来，每次改变某一位数，判断如果是素数入队，否则舍弃。

#include <cstdio>#include <iostream>#include <cstring>#include <map>#include <queue>#include <cmath>using namespace std;const int N = 1e4 + 10;bool is[N];int first, last;map <int, int> prime;int vis[N];struct node {int num, step;};void getPrime() {int i, j, s, e = sqrt(N);memset(is, true, sizeof(is));prime.clear();is[0] = is[1] = false;for (i = 4; i < N; i += 2) is[i] = false;for (i = 3; i < e; i += 2) if (is[i]) {for (j = i * i, s = i + i; j < N; j += s) {is[j] = false;}}for (i = 1000; i < 10000; i++) {if (is[i]) prime[i] = 1;}}int getD(int a, int b) {for (int i = 0; i < b; i++) {a /= 10;}return a % 10;}int getNum(int a, int b, int c) {for (int i = 0; i < b; i++) {c *= 10;}return a + c;}int bfs() {node tmp = {first, 0};if (first == last) return 0;queue<node> q;q.push(tmp);while (q.empty() != 1) {tmp = q.front();q.pop();node now;now.step = tmp.step + 1;for (int i = 0; i < 4; i++) {int j = i == 3 ? 1 : 0;int digit = getD(tmp.num, i);for (; j < 10; j++) {now.num = getNum(tmp.num, i, j - digit);if (prime[now.num] == 1) {if (now.num == last) return now.step;if (vis[now.num] == 0){q.push(now);vis[now.num] = 1;}}}}}return -1;}int main() {int t;getPrime();scanf("%d", &t);while (t--) {scanf("%d%d", &first, &last);int ans = bfs();memset(vis, 0, sizeof(vis));printf("%d\n", ans);}return 0;}

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