poj1872A Dicey Problem(迭代加深dfs)

A Dicey Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 788		Accepted: 271

Description

The three-by-three array in Figure 1 is a maze. A standard six-sided die is needed to traverse the maze (the layout of a standard six-sided die is shown in Figure 2). Each maze has an initial position and an initial die configuration. In Figure 1, the starting position is row 1, column 2-the "2" in the top row of the maze-and the initial die configuration has the "5" on top of the die and the "1" facing the player (assume the player is viewing the maze from the bottom edge of the figure). 

To move through the maze you must tip the die over on an edge to land on an adjacent square, effecting horizontal or vertical movement from one square to another. However, you can only move onto a square that contains the same number as the number displayed on the top of the die before the move, or onto a "wild" square which contains a star. Movement onto a wild square is always allowed regardless of the number currently displayed on the top of the die. The goal of the maze is to move the die off the starting square and to then find a way back to that same square. 

For example, at the beginning of the maze there are two possible moves. Since the 5 is on top of the die, it is possible to move down one square, and since the square to the left of the starting position is wild it is also possible to move left. If the first move chosen is to move down, this brings the 6 to the top of the die and moves are now possible both to the right and down. If the first move chosen is instead to the left, this brings the 3 to the top of the die and no further moves are possible. 

If we consider maze locations as ordered pairs of row and column numbers (row, column) with row indexes starting at 1 for the top row and increasing toward the bottom, and column indexes starting at 1 for the left column and increasing to the right, the solution to this simple example maze can be specified as: (1,2), (2,2), (2,3), (3,3), (3,2), (3,1), (2,1), (1,1), (1,2). A bit more challenging example maze is shown in Figure 3. 

The goal of this problem is to write a program to solve dice mazes. The input file will contain several mazes for which the program should search for solutions. Each maze will have either a unique solution or no solution at all. That is, each maze in the input may or may not have a solution, but those with a solution are guaranteed to have only one unique solution. For each input maze, either a solution or a message indicating no solution is possible will be sent to the output. 

Input

The input file begins with a line containing a string of no more than 20 non-blank characters that names the first maze. The next line contains six integers delimited by single spaces. These integers are, in order, the number of rows in the maze (an integer from 1 to 10, call this value R), the number of columns in the maze (an integer from 1 to 10, call this value C), the starting row, the starting column, the number that should be on top of the die at the starting position, and finally the number that should be facing you on the die at the starting position. The next R lines contain C integers each, again delimited by single spaces. This R * C array of integers defines the maze. A value of zero indicates an empty location in the maze (such as the two empty squares in the center column of the maze in Figure 3), and a value of -1 indicates a wild square. This input sequence is repeated for each maze in the input. An input line containing only the word "END" (without the quotes) as the name of the maze marks the end of the input.

Output

The output should contain the name of each maze followed by its solution or the string "No Solution Possible" (without the quotes). All lines in the output file except for the maze names should be indented exactly two spaces. Maze names should start in the leftmost column. Solutions should be output as a comma-delimited sequence of the consecutive positions traversed in the solution, starting and ending with the same square (the starting square as specified in the input). Positions should be specified as ordered pairs enclosed in parentheses. The solution should list 9 positions per line (with the exception of the last line of the solution for which there may not be a full 9 positions to list), and no spaces should be present within or between positions.

Sample Input

DICEMAZE1
3 3 1 2 5 1
-1 2 4
5 5 6
6 -1 -1
DICEMAZE2
4 7 2 6 3 6
6 4 6 0 2 6 4
1 2 -1 5 3 6 1
5 3 4 5 6 4 2
4 1 2 0 3 -1 6
DICEMAZE3
3 3 1 1 2 4
2 2 3
4 5 6
-1 -1 -1
END

Sample Output

DICEMAZE1
  (1,2),(2,2),(2,3),(3,3),(3,2),(3,1),(2,1),(1,1),(1,2)
DICEMAZE2
  (2,6),(2,5),(2,4),(2,3),(2,2),(3,2),(4,2),(4,1),(3,1),
  (2,1),(2,2),(2,3),(2,4),(2,5),(1,5),(1,6),(1,7),(2,7),
  (3,7),(4,7),(4,6),(3,6),(2,6)
DICEMAZE3
  No Solution Possible

Source

World Finals 1999

题目大意：给一个地图，n*m的，地图上每个点有相应的数字，-1~6，-1代表空地，0代表障碍。现在给你一个骰子，如果骰子旁边的位置上的数字恰好等于当前骰子状态的上表面的点数或者旁边是空地，则骰子可以滚动到该位置。给你初始骰子的位置以及初始骰子的状态，求能否找到一条路径使骰子走一圈后又回到起点，如果有，按要求输出路径。

题目分析：dfs输出路径问题。由于深度未知，可以用迭代加深dfs。每个骰子有24个状态，可以枚举出来，只要定了骰子的上表面和前表面，骰子的状态其实就已经定了，所以可以用四维数组判重。理论上一共有24*m*n个状态，但是最后经过测试发现其实深度2*m*n也能过。最后再经过测试发现其实不用设置深度也能过。dfs遍历答案是唯一的，题目好像说了如果有解保证是唯一解。

这题输出有点搞人，小心PE，虽然我没有PE。

最后隆重友情鸣谢峰峰、吉吉两位小朋友，帮我找出本题最大的trick：搜索的时候起点不能标记！！如果标记了样例虽然能过，但是有的例子是过不了的，所以会WA，过不了是因为有的测试数据回到起点的状态和初始状态会一样，如果一开始就把起点标记了，就回不到起点了。。。。

详情请见代码：

#include <iostream>
#include<cstdio>
#include<cctype>
#include<cstring>
using namespace std;
const int N = 12;
bool flag[N][N][N][N];//
int map[N][N];
bool ok;
int path[105][2];
char s[25];
//u,d,f,b,l,r   24state
int die[][6] = {
                {1,6,2,5,3,4},{1,6,3,4,5,2},{1,6,5,2,4,3},{1,6,4,3,2,5},
                {2,5,1,6,4,3},{2,5,4,3,6,1},{2,5,6,1,3,4},{2,5,3,4,1,6},
                {3,4,1,6,2,5},{3,4,2,5,6,1},{3,4,6,1,5,2},{3,4,5,2,1,6},
                {4,3,1,6,5,2},{4,3,5,2,6,1},{4,3,6,1,2,5},{4,3,2,5,1,6},
                {5,2,1,6,3,4},{5,2,3,4,6,1},{5,2,6,1,4,3},{5,2,4,3,1,6},
                {6,1,2,5,4,3},{6,1,4,3,5,2},{6,1,5,2,3,4},{6,1,3,4,2,5},
};
int n,m,si,sj,up,fa;
int ctrl;

void dfs(int curi,int curj,int dp,int u,int d,int f,int b,int l,int r)
{
    if(dp > ctrl)
        return;
    if(ok)
        return;
    if(flag[curi][curj][u][f])
        return;
    if(dp)
        flag[curi][curj][u][f] = 1;
    path[dp][0] = curi;
    path[dp][1] = curj;
    if(curi == si && curj == sj && dp)
    {
        int i,j;
        for(i = 0;i <= dp;i ++)
        {
            if(i % 9 == 0)
                printf("  ");
            if((i + 1) % 9)
            {
                printf("(%d,%d)",path[i][0],path[i][1]);
                if(i < dp)
                    putchar(',');
            }
            else
            {
                printf("(%d,%d)",path[i][0],path[i][1]);
                if(i < dp)
                    putchar(',');
                putchar(10);
            }
        }
        if((dp + 1) % 9)
            putchar(10);
        ok = true;
        return;
    }
//    printf("%d  %d\n",curi,curj);
//    printf("%d  %d  %d  %d  %d  %d\n",u,d,f,b,l,r);
//    system("pause");
    if(curi > 1 && (map[curi - 1][curj] == u || map[curi - 1][curj] == -1))
        dfs(curi - 1,curj,dp + 1,f,b,d,u,l,r);
    if(curi < n && (map[curi + 1][curj] == u || map[curi + 1][curj] == -1))
        dfs(curi + 1,curj,dp + 1,b,f,u,d,l,r);
    if(curj > 1 && (map[curi][curj - 1] == u || map[curi][curj - 1] == -1))
        dfs(curi,curj - 1,dp + 1,l,r,f,b,d,u);
    if(curj < m && (map[curi][curj + 1] == u || map[curi][curj + 1] == -1))
        dfs(curi,curj + 1,dp + 1,r,l,f,b,u,d);

    flag[curi][curj][u][f] = 0;
}

int nextint()
{
    char c;
    int ret;
    bool sig = false;
    while(isspace(c = getchar()))
        ;
    if(c == '-')
    {
        sig = true;
        c = getchar();
    }
    ret = c - '0';
    while((c = getchar()) >= '0' && c <= '9')
        ret = ret * 10 + c - '0';
    return sig?-ret:ret;
}

int main()
{
    int i,j;
    while(~scanf("%s",s))
    {
        if(strcmp(s,"END") == 0)
            break;
        n = nextint();
        m = nextint();
        si = nextint();
        sj = nextint();
        up = nextint();
        fa = nextint();
        for(i = 1;i <= n;i ++)
            for(j = 1;j <= m;j ++)
                map[i][j] = nextint();
        memset(flag,0,sizeof(flag));
        ok = false;
        printf("%s\n",s);
        for(i = 0;i < 24;i ++)
            if(die[i][0] == up && die[i][2] == fa)
                break;
        //printf("u:%d d:%d f:%d b:%d l:%d r:%d\n",die[i][0],die[i][1],die[i][2],die[i][3],die[i][4],die[i][5]);
        ctrl = 1;
        while(1)
        {
            if(ctrl >= m * n * 6 || ok)
                break;
            dfs(si,sj,0,die[i][0],die[i][1],die[i][2],die[i][3],die[i][4],die[i][5]);
            ctrl ++;
        }
        //printf("%d\n",ctrl);
        if(ok == false)
            printf("  No Solution Possible\n");
    }
    return 0;
}
//156K	47MS

不用迭代加深也能过！