hihocoder 1426 What a Ridiculous Election

题目链接：What a Ridiculous Election

题目大意：有三种操作：
1、交换相邻两个数
2、对某一个数位上的数，数值加一，对十求余，这个操作最多用3次
3、对某一个数位上的数，数值乘二，对十求余，这个操作最多用2次
现在问你能不能从12345变成给定的数字（五位数），能的话输出最少的操作数

题目思路：BFS预处理一下，res数组记一下数字，剩余的操作2，剩余的操作3，然后模拟的时候好好做一下就好了

#include <map>#include <set>#include <cmath>#include <stack>#include <queue>#include <vector>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 1e5+10;const int inf = 0x3f3f3f3f;char s[10];bool flag,vis[10];int res[maxn][5][5];struct node{    int a[10],sec,thr,step;    bool operator < (const node &a)const{        return step > a.step;    }};node target;int check(node now){    int num = 0;    for(int i = 1;i <= 5;i++)        if(now.a[i] != target.a[i]) return 0;    return 1;}int turn(node now){    int sum = 0;    for(int i = 1;i <= 5;i++){        sum += now.a[i];        sum *= 10;    }    sum /= 10;    return sum;}void bfs(node now){    priority_queue<node>q;    q.push(now);    int tmp = turn(now);    res[tmp][now.sec][now.thr] = 0;    node u,next;    bool flag;    while(!q.empty()){        u = q.top();        q.pop();        for(int i = 2;i <= 5;i++){            next = u;            next.step++;            swap(next.a[i],next.a[i-1]);            int p = turn(next);            if(next.step >= res[p][next.sec][next.thr]) continue;            q.push(next);            res[p][next.sec][next.thr] = next.step;        }        if(u.sec > 0){            for(int i = 1;i <= 5;i++){                next = u;                next.sec--;                next.step++;                next.a[i] = (next.a[i]+1)%10;                int p = turn(next);                if(next.step >= res[p][next.sec][next.thr]) continue;                q.push(next);                res[p][next.sec][next.thr] = next.step;            }        }        if(u.thr > 0){            for(int i = 1;i <= 5;i++){                next = u;                next.thr--;                next.step++;                next.a[i] = (next.a[i]*2)%10;                int p = turn(next);                if(next.step >= res[p][next.sec][next.thr]) continue;                q.push(next);                res[p][next.sec][next.thr] = next.step;            }        }    }}int main(){    node now;    now.sec = 3,now.thr = 2,now.step = 0;    for(int i = 1;i <= 5;i++) now.a[i] = i;    memset(res,inf,sizeof(res));    bfs(now);    while(~scanf("%s",s)){        for(int i = 1;i <= 5;i++) target.a[i] = s[i-1]-'0';        int p = turn(target);        int ans = inf;        for(int i = 0;i <= 3;i++)            for(int j = 0;j <= 2;j++)                ans = min(ans,res[p][i][j]);        if(ans == inf) puts("-1");        else printf("%d\n",ans);    }    return 0;}