UVA 225 - Golygons（dfs回溯）

题意：

一个探险队要去参观城市，走n次，每次走1,2,3,4,5,…n步，参观停留的城市，每次方向转90度。路途中有一些障碍物是不能走的，之前拐弯过的点也是不能走的。求所有走法，按字典序输出。

解析：

直接回溯深搜就可以了。

AC代码

[code]#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
using namespace std;
const int N = 1000;
const int LEN = 500;
const char str[] = "nsew";
//  N  S  W  E
const int dx[] = { 1,-1, 0, 0};
const int dy[] = { 0, 0, 1,-1};
string ans
;
int last, m, tot, path
;
bool grid

, vis

;

bool can_move(int x, int y, int xx, int yy) {
    int from, to;
    if(x == xx) {
        from = min(y, yy);
        to = max(y, yy);
        for(int i = from; i <= to; i++)
            if(grid[xx][i]) return false;
    }else if(y == yy) {
        from = min(x, xx);
        to = max(x, xx);
        for(int i = from; i <= to; i++)
            if(grid[i][yy]) return false;
    }
    return true;
}
void dfs(int x, int y, int dir, int cur) {
    if(cur == last) {
        if(x == LEN && y == LEN) {
            ans[tot] = "";
            for(int i = 0; i < last; i++)
                ans[tot] += str[path[i]];
            tot++;
        }
        return ;
    }
    int xx, yy, d;
    if(dir == 0 || dir == 1) d = 2;
    else d = 0;
    for(int i = 0; i < 2; i++) {
        xx = x + (cur+1)*dx[d+i];
        yy = y + (cur+1)*dy[d+i];
        if(!vis[xx][yy] && can_move(x, y, xx, yy)) {
            path[cur] = d+i;
            vis[xx][yy] = true;
            dfs(xx, yy, d+i, cur+1);
            vis[xx][yy] = false;
        }
    }
}
int main() {
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    while(T--) {
        memset(vis, 0, sizeof(vis));
        memset(grid, 0, sizeof(grid));
        scanf("%d%d",&last, &m);
        int x, y;
        for(int i = 0; i < m; i++) {
            scanf("%d%d", &x, &y);
            grid[y+LEN][x+LEN] = true;
        }
        tot = 0;
        for(int i = 0; i < 4; i++) {
            memset(vis, false, sizeof(vis));
            path[0] = i;
            dfs(LEN+dx[i], LEN+dy[i], i, 1);
        }
        sort(ans, ans+tot);
        for(int i = 0; i < tot; i++) {
            puts(ans[i].c_str());
        }
        printf("Found %d golygon(s).\n\n", tot);
    }
    return 0;
}