CSU~1336: Interesting Calculator(spfa)

1336: Interesting Calculator

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Description

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

 

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 2561 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 11 1 1 1 1 1 1 1 1 112 256100 100 100 1 100 100 100 100 100 100100 100 100 100 100 1 100 100 100 100100 100 10 100 100 100 100 100 100 100

Sample Output

Case 1: 2 2Case 2: 12 3

Hint

Source

湖南省第九届大学生计算机程序设计竞赛

刚一看果断就想要广搜，不过听说广搜会超时，看了大佬的代码，居然能用spfa写，感觉图论学X身上了

由于数据只有100000，所以可以开数组，用spfa直接跑一遍就行了，简单粗暴，却又令人反思

#include <iostream>#include<string>#include<string.h>#include<stdio.h>#include<stdlib.h>#include<queue>#include<math.h>#include<algorithm>#define ll long long#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define maxn 100000+5using namespace std;int c[3][10],st,ed;int dis[maxn],num[maxn];int vis[maxn];int Case=1;void spfa(){    mem(dis,inf);    mem(vis,0);    queue<int>q;    int u,v,t;    q.push(st);    vis[st]=1;    num[st]=dis[st]=0;    while(!q.empty())    {        u=q.front();        q.pop();        vis[u]=0;        for(int i=0;i<3;i++)        {            for(int j=0;j<10;j++)            {                if(i==0)                v=u*10+j;                else if(i==1)                v=u+j;                else                v=u*j;                if(v>ed)continue;                if(dis[v]>dis[u]+c[i][j])                {                    dis[v]=dis[u]+c[i][j];                    num[v]=num[u]+1;                    if(!vis[v])                    {                        vis[v]=1;                        q.push(v);                    }                }            }        }    }      printf("Case %d: %d %d\n",Case++,dis[ed],num[ed]);}int main(){    while(~scanf("%d%d",&st,&ed))    {        for(int i=0;i<3;i++)        {            for(int j=0;j<10;j++)            {                scanf("%d",&c[i][j]);            }        }        spfa();    }    return 0;}