uva 10401 Injured Queen Problem(DP)

Problem I
Injured Queen Problem
Input: standard input
Output: standard output
Time Limit: 6 seconds
Memory Limit: 32 MB

Chess is a two-player board game believed to have been played in India as early as the sixth century. However, in this problem we will not discuss about chess, rather we will talk about a modified form of the classic n-queen problem. I know you are familiar with plotting n-queens on a chess board with the help of a classic backtracking algorithm. If you write that algorithm now you will find that there are92 ways of plotting 8 queens in an 8x8 board provided no queens attack each other.

In this problem we will talk about injured queens who can move only like a king in horizontal and diagonal direction from current position but can reach any row from current position like a normal chess queen. You will have to find the number of possible arrangements with such injured queens in a particular (n x n) board (with some additional constraints), such that no two queens attack each other.

Fig: Injured Queen at a6 can reach the adjacent grey squares. Queen ate4 can reach adjacent grey squares. The injured queen positions are black and the reachable places are grey.

Input

Input file contains several lines of input. Each line expresses a certain board status. The length of these status string is the board dimensionn (0<n<=15). The first character of the string denotes the status of first column, the second character of the string denotes the status of the second column and so on. So if the first character of the status string is2, it means that we are looking for arrangements (no two injured queen attack each other) which has injured queen in columna, row 2. The possible numbers for rows are 1, 2, 3… D, E, F which indicates row 1, 2, 3… 13, 14, 15. If any column contains ‘?’ it means that in that column the injured queen can be in any row. So a status string1?4??3 means that you are asked to find out total number of possible arrangements in a(6x6) chessboard which has three of its six injured queens at a1, c4 and f3. Also note that there will be no invalid inputs. For example“1?51” is an invalid input because a (4x4) chessboard does not have a fifth row.

Output

For each line of input produce one line of output. This line should contain an integer which indicates the total number of possible arrangements of the corresponding input status string.

Sample Input
??????
???????????????
???8?????
43?????

Sample Output
2642
22696209911206174
2098208
0

#include <iostream>#include <cstdio>#include <string>#include <cstring>using namespace std;const int maxn = 20;long long int dp[maxn][maxn];string str;int num[maxn];void initial(){for(int i = 0;i < maxn;i++){for(int j = 0;j < maxn;j++){dp[i][j] = 0;}num[i] = -1;}}void computing(){int len = str.length();for(int i = 0;i < len;i++){if(str[i] == '?'){num[i] = -1;}if(str[i] >= '0' && str[i] <= '9'){num[i] = str[i]-'1';}if(str[i] >= 'A' && str[i] <= 'Z'){num[i] = 9+str[i]-'A';}}//cout << num[0] << " " << dp[0][3] <<endl;if(num[0] != -1){dp[0][num[0]] = 1;}else{for(int i = 0;i < len;i++){dp[0][i] = 1;}}//cout << num[0] << " " << dp[0][3] <<endl;for(int i = 1;i < len;i++){if(num[i] == -1){for(int j = 0;j < len;j++){for(int k = j-2;k >= 0;k--){dp[i][j] += dp[i-1][k];}for(int k = j+2;k < len;k++){dp[i][j] += dp[i-1][k];}}}else{for(int k = num[i]-2;k >= 0;k--){dp[i][num[i]] += dp[i-1][k];}for(int k = num[i]+2;k < len;k++){dp[i][num[i]] += dp[i-1][k];}}}long long ans = 0;for(int i = 0;i < len;i++){ans += dp[len-1][i];//cout << dp[len-1][i] << endl;}cout << ans << endl;}int main(){while(cin >> str){initial();computing();}return 0;}