coj1090: Number Transformation bfs

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这道题用bfs做，一开始没理解清题意WA的不要不要的，后来仔细看题发现A是一步变一次的，也就是说每次可以加的素数的范围是改变的，比如当S=5，T=20的时候，S先加3=8，然后此时可以取的素数就不仅有2、3，还有5、7了，一开始不仔细看题真是醉了。

Description

In this problem, you are given a pair of integers A and B. You can transform any integer number A to B by adding x to A.This x is an integer number which is a prime below A.Now,your task is to find the minimum number of transformation required to transform S to another integer number T.

Input

Input contains multiple test cases.Each test case contains a pair of integers S and T(0< S < T <= 1000) , one pair of integers per line.

Output

For each pair of input integers S and T you should output the minimum number of transformation needed as Sample output in one line. If it's impossible ,then print 'No path!' without the quotes.

Sample Input

5 73 4

Sample Output

Need 1 step(s)No path!
bfs注意不要重复放一个数，不然队列就爆了，然后可以将素数先打表，我这里还将每个数的前一个最大素数打表了，用的时候有种链表的快感
下面是AC代码
#include<iostream>#include<cstdio>#include<queue>#include<cstring>using namespace std;int preprime[1005], prime[1005] = { 0 };void bfs(int S,int T,int step[], int vis[]){int temp1, ts;queue<int>p;step[S] = 1;p.push(S);while (!p.empty()){temp1 = p.front();ts=temp1;p.pop();while (preprime[ts] != 0){if (temp1 + preprime[ts]<T){if (vis[temp1 + preprime[ts]] == 0){p.push(temp1 + preprime[ts]);  step[temp1 + preprime[ts]] = step[temp1] + 1;vis[temp1 + preprime[ts]]=1;}}else if (temp1 + preprime[ts] == T){cout << "Need " << step[temp1] << " step(s)" << endl;return;}ts = preprime[ts];}}cout << "No path!" << endl;}int main(){int S, T;int step[1005] = {0};int vis[1005] = {0};int temp = 0;for (int i = 2; i<502; i++)for (int j = 2 * i; j<1002; j += i)prime[j] = 1;for (int i = 2; i<1002; i++){preprime[i] = temp;if (prime[i] == 0)temp = i;}while (scanf("%d%d",&S,&T)==2){if(S==1||S==2)cout << "No path!" << endl;else{for(int i=0;i<T;i++){vis[i]=0;step[i]=0;}  bfs(S,T,step, vis);}}return 0;}

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