POJ3126 Prime Path 素数

         这题先将10000以内的素数都找出来，然后在满足条件的每对素数连上一个权为1的边，接着用dijkstra计算一下最短路就可以。

#ifndef HEAD#include <stdio.h>#include <vector>#include <math.h>#include <string.h>#include <string>#include <iostream>#include <queue>#include <list>#include <algorithm>#include <stack>#include <map>using namespace std;#endif // !HEAD#ifndef QUADMEMSETinline void QuadMemSet(void* dst, int iSize, int value){iSize = iSize / 4;int* newDst = (int*)dst;#ifdef WIN32__asm{mov edi, dstmov ecx, iSizemov eax, valuerep stosd}#elsefor (int i = 0; i < iSize; i++){newDst[i] = value;}#endif}#endifstruct EDGE{int from;int to;int cost;EDGE* next;};int Dij(int s, int to, vector<EDGE*> &head,int N){vector<int> visited;visited.resize(N, 0);vector<int> res;res.resize(N, 10000000);res[s] = 0;//int t = s;while (true){int v = -1;for (int t = 0; t<N; t++){if (visited[t] == 0 && (v == -1 || res[v] > res[t])){v = t;}}if (v == -1 || res[v] >= 10000000){break;}visited[v] = 1;for (EDGE* p = head[v]; p; p = p->next){if (visited[p->to] == 0){res[p->to] = min(res[p->to], res[v] + p->cost);}}}return res[to];}EDGE edges[20000];int main(){int is_prime[101];int ab_isprime[10000];memset(is_prime, 1, sizeof(is_prime));memset(ab_isprime, 1, sizeof(ab_isprime));is_prime[0] = is_prime[1] = 0;for (int i = 2; i * i < 10001;i++){if (is_prime[i]){for (int j = 2 * i; j < 101;j += i){is_prime[j] = 0;}for (int j = max(2, (1000 + i - 1) / i) * i; j < 10000; j += i){ab_isprime[j] = 0;}}}vector<int> abPrimes;map<int, int> mapIndex;for (int i = 1000; i < 10000;i++){if (ab_isprime[i]){abPrimes.push_back(i);mapIndex[i] = abPrimes.size() - 1;}}vector<EDGE*> head;head.resize(abPrimes.size(), NULL);int count = 0;for (int i = 0; i < abPrimes.size();i++){for (int j = 0; j < abPrimes.size();j++){int countofzero = 0;int div = 1;while (div <= 1000){if ((abPrimes[i]/div) % 10 - abPrimes[j] / div % 10 == 0){countofzero++;}div *= 10;}if (i == j){continue;}if (countofzero == 3){EDGE edge;edge.from = i;edge.to = j;edge.cost = 1;edge.next = head[i];EDGE invedge = edge;invedge.from = j;invedge.to = i;edges[count] = edge;;head[i] = &edges[count++];invedge.next = head[j];edges[count] = invedge;head[j] = &edges[count++];}}}#ifdef _DEBUGfreopen("d:\\in.txt", "r", stdin);#endifint N;scanf("%d\n", &N);for (int i = 0; i < N;i++){int a, b;scanf("%d %d\n", &a, &b);if (a == b){printf("0\n");}else{printf("%d\n", Dij(mapIndex[a], mapIndex[b], head, abPrimes.size()));}}return 0;}