poj-3126

来源：互联网发布：java http multi 编辑：程序博客网时间：2024/05/17 21:27

// 600K 16MS    G++#include <cstdio>#include <queue>#include <cstring>#include <cmath>using namespace std;char compositeNum[10000];char BFSFlag[10000];void getPrime() {    int max = 100;    int begin = 2;    while(begin <= 100) {        for (int i = begin + 1; i <= 9999; i++) {            if (i%begin == 0) {                compositeNum[i] = 1;            }        }        for (int i = begin + 1; i <= 999; i++) {            if (compositeNum[i] == 0) {                begin = i;                break;            }        }    }}int sNum;int tNum;struct BFSNode {    int num;    int roundNum;}; typedef struct BFSNode BFSNode;queue<BFSNode> BFSQueue;int BFS() {    while(BFSQueue.size()) {        BFSQueue.pop();    }    memset(BFSFlag, 0, sizeof(BFSFlag));    BFSNode beignNode;    beignNode.num = sNum;    beignNode.roundNum = 0;    BFSQueue.push(beignNode);    BFSFlag[sNum] = 1;    while(BFSQueue.size()) {        BFSNode curNode = BFSQueue.front();        BFSQueue.pop();        int curNum = curNode.num;        int curRoundNum = curNode.roundNum;        //1        for (int i = 1; i <= 9; i++) {            int newVal = i*1000 + curNum%1000;            if (!compositeNum[newVal]) {                if (newVal == tNum) {                    return curRoundNum+1;                } else {                    if (!BFSFlag[newVal]) {                        BFSFlag[newVal] = 1;                        BFSNode newNode;                        newNode.num = newVal;                        newNode.roundNum = curRoundNum + 1;                        BFSQueue.push(newNode);                    }                }            }        }        //2        for (int i = 0; i <= 9; i++) {            int newVal = curNum/1000 * 1000 + i*100 + curNum%100;            if (!compositeNum[newVal]) {                if (newVal == tNum) {                    return curRoundNum+1;                } else {                    if (!BFSFlag[newVal]) {                        BFSFlag[newVal] = 1;                        BFSNode newNode;                        newNode.num = newVal;                        newNode.roundNum = curRoundNum + 1;                        BFSQueue.push(newNode);                    }                }            }        }        //3        for (int i = 0; i <= 9; i++) {            int newVal = curNum/100 * 100 + i*10 + curNum %10;            if (!compositeNum[newVal]) {                if (newVal == tNum) {                    return curRoundNum+1;                } else {                    if (!BFSFlag[newVal]) {                        BFSFlag[newVal] = 1;                        BFSNode newNode;                        newNode.num = newVal;                        newNode.roundNum = curRoundNum + 1;                        BFSQueue.push(newNode);                    }                }            }        }        //4        for (int i = 0; i <= 9; i++) {            int newVal = curNum/10*10 + i;            if (!compositeNum[newVal]) {                if (newVal == tNum) {                    return curRoundNum+1;                } else {                    if (!BFSFlag[newVal]) {                        BFSFlag[newVal] = 1;                        BFSNode newNode;                        newNode.num = newVal;                        newNode.roundNum = curRoundNum + 1;                        BFSQueue.push(newNode);                    }                }            }        }    }    return -1;}void solve() {    printf("%d\n", BFS());}int main() {    memset(compositeNum, 0, sizeof(compositeNum));    getPrime();    // for (int i = 9000; i < 10000; i++) {    //     if (!compositeNum[i]) {    //         printf("%d\n", i);    //     }    // }    int caseNum;    scanf("%d\n", &caseNum);    for (int i = 1; i <= caseNum; i++) {        scanf("%d %d", &sNum, &tNum);        if (sNum == tNum) {            printf("0\n");        } else {            solve();        }    }}

算是一道数论 + BFS水题。

解题之前先用素数筛子筛一遍1~10000的数，搞一个bitmap数组标示素数和非素数。

然后就简单了，BFS从给的数开始，每次有9（第一位不能是0）*10*10*10个选择，不过因为有BFSFlag标示和可能不是素数，因此其实最后的运算量不大，

最差就是把1000~9999的数全过一遍，以遇到要转的素数结束BFS。BFSNode带上round次数，最后输出。

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