【dfs/bfs+记录路径】Transformation: from A to B

Think:
1知识点：dfs/bfs+记录路径
2题意：输入两个数a,b，询问通过两种操作是否可以由a得到b
操作1:a = a*2
操作2:a = a*10 + 1

vjudge题目链接

以下为Accepted代码——bfs+记录路径

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;struct Node{    LL num;    int size;}link[104014];int op, tp;int pre[104014];void bfs(LL a, LL b);void pri_path(int id);int main(){    LL a, b;    while(~scanf("%lld %lld", &a, &b)){        if(a == b){            printf("YES\n");            printf("1\n");            printf("%lld\n", a);        }        else bfs(a, b);    }    return 0;}void bfs(LL a, LL b){    int id = -1;    Node t, tmp;    op = tp = 0;    memset(pre, -1, sizeof(pre));    link[tp].num = a, link[tp].size = 1;    tp++;    while(op < tp){        t = link[op++];        tmp.num = t.num << 1;        tmp.size = t.size + 1;        if(tmp.num < b){            link[tp++] = tmp;            pre[tp-1] = op-1;        }        else if(tmp.num == b){            link[tp++] = tmp;            pre[tp-1] = op-1;            id = tp-1;            break;        }        tmp.num = t.num * (LL)10 + (LL)1;        tmp.size = t.size + 1;        if(tmp.num < b){            link[tp++] = tmp;            pre[tp-1] = op-1;        }        else if(tmp.num == b){            link[tp++] = tmp;            pre[tp-1] = op-1;            id = tp-1;            break;        }    }    if(id == -1) {        printf("NO\n");    }    else {        printf("YES\n");        printf("%d\n", link[id].size);        pri_path(id);        printf("\n");    }    return;}void pri_path(int id){    if(pre[id] != -1)        pri_path(pre[id]);    if(pre[id] == -1) printf("%lld", link[id].num);    else printf(" %lld", link[id].num);}

以下为Accepted代码——dfs

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long LL;bool flag;LL a, b;int tp;LL link[1014];void dfs(LL x);int main(){    while(~scanf("%lld %lld", &a, &b)){        tp = 0, flag = false;        link[tp++] = a;        dfs(a);        if(flag) {            printf("YES\n");            printf("%d\n", tp);            for(int i = 0; i < tp; i++)                printf("%lld%c", link[i], i == tp-1? '\n': ' ');        }        else {            printf("NO\n");        }    }    return 0;}void dfs(LL x){    if(flag || x > b) return;    link[tp++] = x<<1;    if(link[tp-1] == b){        flag = true;        return;    }    dfs(link[tp-1]);    if(flag) return;    else tp--;    link[tp++] = x*(LL)10 + (LL)1;    if(link[tp-1] == b){        flag = true;        return;    }    dfs(link[tp-1]);    if(flag) return;    else tp--;    return;}