UVA10384-The Wall Pushers（迭代加深搜索）

Problem UVA10384-The Wall Pushers

Accept: 199   Submit: 1546
Time Limit: 10000 mSec

Problem Description

Input

The input file may contain several mazes to solve. Each maze description starts with a single line containing two integers x and y (1 ≤ x ≤ 6,1 ≤ y ≤ 4) which is the start position in the maze. Next follows four lines with six integers each. These integers p (0 ≤ p ≤ 15) describe each square in the maze in the following way: p is the sum of 1 (if there is a wall west of the square), 2 (north), 4 (east) and 8 (south). Each inner wall will thus be mentioned twice. Each opening in the boundary is considered an exit. The input ends with a maze with starting coordinates 0, 0 and should not be processed.

 Output

Output a single line for each maze with the description of a path with minimum length that leads to any of exits. Use the letters ‘N’, ‘S’, ‘E’ and ‘W’ to denote north, south, east and west, respectively. If there are several solutions with minimum length, display any one of them.

 

 Sample Input

2 3
10 2 10 10 2 6
3 12 11 14 9 4
13 15 3 6 15 13
14 11 12 9 14 11
0 0

 

 Sample Output

NESESEENNWNWWWWW

题解：这个题最大的提示在Input里，直接告诉你最佳存图方式，最短路径，因此采用IDA*，之后就是水题了。

#include <bits/stdc++.h>

using namespace std;

const int maxn = 5, maxm = 7;
const char direction[] = "WNES";
const int INF = 0x3f3f3f3f;
const int dx[] = { 0,-1,0,1 };
const int dy[] = { -1,0,1,0 };
const int dir[] = { 1,2,4,8 };

int sx, sy, maxd;
int ans[maxn*maxm];
int gra[maxn][maxm];
bool vis[maxn][maxm];

vector< pair<int, int> > Exit;

int ok(int x,int y) {
    if (x == 1 && !(gra[x][y] & 2)) return 1;
    else if (x == 4 && !(gra[x][y] & 8)) return 3;
    if (y == 1 && !(gra[x][y] & 1)) return 0;
    else if (y == 6 && !(gra[x][y] & 4)) return 2;
    return -1;
}

bool Judge(int x, int y) {
    if (x < 1 || y < 1 || x > 4 || y > 6) return true;
    return false;
}

bool dfs(int d, int x, int y) {
    if (d == maxd) return false;
    int tmp = ok(x, y);
    if (tmp != -1) {
        ans[d] = tmp;
        return true;
    }

    int h = INF;
    for (vector< pair<int, int> >::iterator it = Exit.begin(); it != Exit.end(); it++) {
        h = min(h, abs(it->first - x) + (it->second - y));
    }
    if (d + h > maxd) return false;

    for (int i = 0; i < 4; i++) {
        int xx = x + dx[i], yy = y + dy[i];
        if (Judge(xx, yy) || vis[xx][yy]) continue;
        if (!(gra[x][y] & dir[i])) {
            vis[xx][yy] = true;
            ans[d] = i;
            if (dfs(d + 1, xx, yy)) return true;
            vis[xx][yy] = false;
        }
        else if (!(gra[xx][yy] & dir[i])) {
            gra[xx][yy] += dir[i];
            gra[x][y] -= dir[i];
            if (!Judge(xx + dx[i], yy + dy[i])) {
                gra[xx + dx[i]][yy + dy[i]] += dir[(i + 2) % 4];
            }
            ans[d] = i;
            vis[xx][yy] = true;
            if (dfs(d + 1, xx, yy)) return true;
            vis[xx][yy] = false;
            if (!Judge(xx + dx[i], yy + dy[i])) {
                gra[xx + dx[i]][yy + dy[i]] -= dir[(i + 2) % 4];
            }
            gra[xx][yy] -= dir[i];
            gra[x][y] += dir[i];
        }
    }
    return false;
}

int main()
{
    //freopen("input.txt", "r", stdin);
    while (~scanf("%d%d", &sx, &sy) && (sx || sy)) {
        for (int i = 1; i <= 4; i++) {
            for (int j = 1; j <= 6; j++) {
                scanf("%d", &gra[i][j]);
                
                if (j == 1) {
                    if (gra[i][j] & 1) Exit.push_back(make_pair(i, j));
                }
                else if (j == 6) {
                    if (gra[i][j] & 1 << 2) Exit.push_back(make_pair(i, j));
                }
            }
            if (i == 1) {
                for (int j = 1; j <= 6; j++) {
                    if (gra[i][j] & (1 << 1)) Exit.push_back(make_pair(i, j));
                }
            }
            else if (i == 4) {
                for (int j = 1; j <= 6; j++) {
                    if (gra[i][j] & (1 << 3)) Exit.push_back(make_pair(i, j));
                }
            }
        }

        for (maxd = 0;; maxd++) {
            memset(vis, false, sizeof(vis));
            vis[sy][sx] = true;
            if (dfs(0, sy, sx)) break;
        }

        for (int i = 0; i < maxd; ++i) {
            printf("%c", direction[ans[i]]);
        }
        printf("
");
    }
}