Triangle Array Problem

描述

The following is a triangle array with 45 numbers from 1, 2, 3, …, 9.

In the above array we find the road 139654278 from A to the bottom. Now your job is find such a
road from A to the bottom.

输入

The input contains several cases. On the first line is a integer n, which is the triangle arrays. Then
follows the descriptions of the n cases. the data of each case consists of 9 lines. On the i-th line contains i symbols from 1 to 9. There is a blank line between two descriptions.

输出

For each triangle, determine if you can find the road from A to the bottom. The road must contain
the 9 symbols from 1 to 9, and they are different. You are not allowed to return to the above line. First
you print “Case #:”, where # is the number of the case. If you can not find such road you print
“Impossible”, otherwise print “Possible”.

样例输入

2
1
3 6
9 7 8
2 6 3 9
3 1 5 8 5
5 9 4 6 3 2
8 3 2 9 4 9 1
6 5 7 4 1 2 8 7
7 1 4 8 7 4 5 6 2

4
3 6
9 7 8
2 6 3 9
3 1 5 8 5
5 9 4 6 3 2
8 3 2 9 4 9 1
6 5 7 4 1 2 8 7
7 1 4 8 7 4 5 6 2

样例输出

Case 1:Possible
Case 2:Impossible

#include<iostream>#include<stdio.h>using namespace std;int cp[9][9], sam[9]={1,3,9,6,5,4,2,7,8};bool flag;void dfs(int x,int y,int k){    if(x==9)    {        flag=true;        return ;    }    if(cp[x][y]==sam[k])    {        if(cp[x+1][y]==sam[k+1]) dfs(x+1,y,k+1);        if(cp[x+1][y+1]==sam[k+1] && !flag) dfs(x+1,y+1,k+1);    }}int main(){    int ca, i, j, k;    scanf("%d",&ca);    k=1;    while(k<=ca)    {        for(i=0;i<9;i++)         for(j=0;j<=i;j++)         cin>>cp[i][j];         flag=false;         dfs(0,0,0);         cout<<"Case "<<k<<":";         if(flag) cout<<"Possible"<<endl;         else cout<<"Impossible"<<endl;         k++;    }    return 0;}