2016 ACM-ICPC 亚洲区域赛北京站E题 What a Ridiculous Election (BFS预处理）

Problem E. What a Ridiculous Election

Description

In country Light Tower, a presidential election is going on. There are two candidates,  Mr. X₁ and Mr. X₂, and both of them are not like good persons. One is called a liar and the other is called a maniac. They tear(Chinese English word, means defame) each other on TV face to face, on newspaper, on internet.......on all kinds of media. The country is tore into two parts because the people who support X₁ are almost as many as the people who support X₂.

After the election day, X₁ and X₂ get almost the same number of votes. No one gets enough votes to win. According to the law of the country, the Great Judge must decide who will be the president. But the judge doesn't want to offend half population of the country, so he randomly chooses a 6 years old kid Tom and authorize him to pick the president. Sounds weird? But the democracy in Light Tower is just like that.

The poor or lucky little kid Tom doesn't understand what is happening to his country. But he has his way to do his job. Tom's ao shu(Chinese English word, means some kind of weird math for kids) teacher just left him a puzzle a few days ago, Tom decide that he who solve that puzzle in a better way will be president. The ao shu teacher's puzzle is like this:

Given a string which consists of five digits('0'-'9'), like "02943", you should change "12345" into it by as few as possible operations. There are 3 kinds of operations:

1. Swap two adjacent digits.

2. Increase a digit by one. If the result exceed 9, change it to itmodulo 10.

3. Double a digit. If the result exceed 9, change it to itmodulo 10.

You can use operation 2 at most three times, and use operation 3 at most twice.

As a melon eater(Chinese English again, means bystander), which candidate do you support? Please help him solve the puzzle.

Input

There are no more than 100,000 test cases.

Each test case is a string which consists of 5 digits.

Output

For each case, print the minimum number of operations must be used to change "12345" into the given string. If there is no solution, print -1.

Sample Input

12435

99999

12374

Sample Output

1

-1

3

【题意】

给出 一个串 问 最少经过多少次变化  使12345 变成这个串；

给出变换方式

1 : 交换相邻两项，  无限制次数

2 ：  某一位增加一    限制 3次  超过10  %10

3： 某一位 *2   限制2次    超过10  %10

【思路】

用一个三维数字 ans【num】【op2】【op3】  预处理 代表  数字为num的  操作2 操作3  用的次数  进行BFS 预处理

【代码】

#include <iostream>#include <stdio.h>#include <algorithm>#include <cmath>#include <math.h>#include <cstring>#include <string>#include <queue>#include <stack>#include <stdlib.h>#include <list>#include <map>#include <set>#include <bitset>#include <vector>#define mem(a,b) memset(a,b,sizeof(a))#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define FIN      freopen("input.txt","r",stdin)#define FOUT     freopen("output.txt", What a Ridiculous Election"w",stdout)#define S1(n)    scanf("%d",&n)#define SL1(n)   scanf("%I64d",&n)#define S2(n,m)  scanf("%d%d",&n,&m)#define SL2(n,m)  scanf("%I64d%I64d",&n,&m)#define Pr(n)     printf("%d\n",n)#define lson rt << 1, l, mid#define rson rt << 1|1, mid + 1, rusing namespace std;typedef long long ll;const double PI=acos(-1);const int INF=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e5+5;const int MOD=1000000007;const int mod=1e9+7;int dir[5][2]={0,1,0,-1,1,0,-1,0};int ans[maxn][5][5];//次数struct node{    int num[10];    int op2,op3;    int step;};int Q_sum(node A){    int sum=0;    for(int i=1;i<=5;i++)    {        sum+=A.num[i];        sum*=10;    }    return sum/10;}void bfs(node t){    queue<node> Q;    mem(ans,INF);    t.op2=3;//  +1    t.op3=2;//  *2    t.step=0;    Q.push(t);    int n=Q_sum(t);    ans[n][t.op2][t.op3]=0;    while(!Q.empty())    {        node u=Q.front();        Q.pop();        for(int i=2;i<=5;i++)//swap        {            node tu=u;            swap(tu.num[i],tu.num[i-1]);            int  num=Q_sum(tu);            tu.step++;            if(tu.step>=ans[num][tu.op2][tu.op3])                continue;            Q.push(tu);            ans[num][tu.op2][tu.op3]=tu.step;        }        if(u.op2>0)//+1        {            for(int i=1;i<=5;i++)            {                node tu=u;                tu.op2--;                tu.num[i]=(tu.num[i]+1)%10;                int num=Q_sum(tu);                tu.step++;                if(tu.step>=ans[num][tu.op2][tu.op3])                    continue;                Q.push(tu);                ans[num][tu.op2][tu.op3]=tu.step;            }        }        if(u.op3>0)//*2        {            for(int i=1;i<=5;i++)            {                node tu=u;                tu.op3--;                tu.num[i]=(tu.num[i]*2)%10;                int num=Q_sum(tu);                tu.step++;                if(tu.step>=ans[num][tu.op2][tu.op3])                    continue;                Q.push(tu);                ans[num][tu.op2][tu.op3]=tu.step;            }        }    }}int main(){    node temp;    for(int i=1;i<=5;i++)        temp.num[i]=i;    bfs(temp);    char str[12];    while(~scanf("%s",str+1))    {        node b;        for(int i=1;i<=5;i++)            b.num[i]=str[i]-'0';        int n=Q_sum(b);        int res=INF;        for(int i=0;i<=3;i++)            for(int j=0;j<=2;j++)        {            res=min(res,ans[n][i][j]);        }        if(res==INF)            printf("-1\n");        else            printf("%d\n",res);    }    return 0;}