The Ninth Hunan Collegiate Programming Contest (2013) Problem I

Problem I

Interesting Calculator

There is an interesting calculator. It has 3 rows of button.

Row 1: button 0, 1, 2, 3, ..., 9. Pressing each button appends that digit to the end of the display.

Row 2: button +0, +1, +2, +3, ..., +9. Pressing each button adds that digit to the display.

Row 3: button *0, *1, *2, *3, ..., *9. Pressing each button multiplies that digit to the display.

Note that it never displays leading zeros, so if the current display is 0, pressing 5 makes it 5 instead of 05. If the current display is 12, you can press button 3, +5, *2 to get 256. Similarly, to change the display from 0 to 1, you can press 1 or +1 (but not both!).

Each button has a positive cost, your task is to change the display from x to y with minimum cost. If there are multiple ways to do so, the number of presses should be minimized.

Input

There will be at most 30 test cases. The first line of each test case contains two integers x and y(0<=x<=y<=10⁵). Each of the 3 lines contains 10 positive integers (not greater than 10⁵), i.e. the costs of each button.

Output

For each test case, print the minimal cost and the number of presses.

Sample Input

12 256
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
12 256
100 100 100 1 100 100 100 100 100 100
100 100 100 100 100 1 100 100 100 100
100 100 10 100 100 100 100 100 100 100

Output for the Sample Input

Case 1: 2 2
Case 2: 12 3

---

The Ninth Hunan Collegiate Programming Contest (2013)
Problemsetter: Rujia Liu
Special Thanks: Feng Chen, Md. Mahbubul Hasan

  典型的广搜，有最优性，用堆来优化可以加速运行，这种试题本质很老，需要认识本质，找准突破口。

#include <iostream>
#include <stdio.h>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <vector>
#include <queue>
#include <set>
#include <algorithm>
#include <map>
#include <stack>
#include <math.h>
#define Max(a,b) ((a)>(b)?(a):(b))
#define Min(a,b) ((a)<(b)?(a):(b))
using namespace std ;
typedef long long LL ;
const LL inf=(LL)10000000000000 ;
int num[4][10] ;
LL dist[100008] ;
LL step[100008] ;
int x, y;
struct  Node{
     int Num ;
     LL money ;
     LL Step ;
     Node(){} ;
     Node(int n ,LL m ,LL s):Num(n),money(m),Step(s){} ;
     friend bool operator <(const Node A ,const Node B){
          if(A.money==B.money)
                return A.Step>B.Step ;
          else
                return A.money>B.money ;
     }
};
void bfs(){
    priority_queue<Node>que ;
    fill(step,step+y+1,inf) ;
    fill(dist,dist+y+1,inf) ;
    que.push(Node(x,0,0)) ;
    dist[x]=0 ;
    step[x]=0 ;
    while(!que.empty()){
        Node now=que.top() ;
        que.pop() ;
        if(now.Num==y){
             return ;
        }
        int next_id ;
        for(int k=1;k<=3;k++){

            for(int i=0;i<=9;i++){
                if(k==1)
                   next_id=now.Num*10+i ;
                else if(k==2)
                   next_id=now.Num+i ;
                else if(k==3)
                   next_id=now.Num*i  ;
                if(next_id>y)
                    continue  ;
                if(now.money+num[k][i]<dist[next_id]){
                       dist[next_id]=now.money+num[k][i] ;
                       step[next_id]=now.Step+1 ;
                       que.push(Node(next_id,dist[next_id],step[next_id])) ;
                }
                else if(now.money+num[k][i]==dist[next_id]){
                       if(now.Step+1<step[next_id]){
                             step[next_id]=now.Step+1 ;
                             que.push(Node(next_id,dist[next_id],step[next_id])) ;
                       }
                }
            }

        }

    }
}
int main(){
    int k=1 ;
    while(scanf("%d%d",&x,&y)!=EOF){
         for(int i=1;i<=3;i++)
            for(int j=0;j<=9;j++)
                scanf("%d",&num[i][j]) ;
         bfs() ;
         printf("Case %d: ",k++) ;
         cout<<dist[y]<<" "<<step[y]<<endl ;
    }
    return 0 ;
}