SCU 4445 Right turn（dfs）题解

思路：离散化之后，直接模拟就行，标记vis开三维

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<stdio.h>
#include<string.h>
#include<queue>
#include<cmath>
#include<map>
#include<set>
#include<vector>
using namespace std;
typedef long long ll;
const int maxn = 1e3 + 10;
const ll mod = 1e8 + 7;
struct node{
    ll x, y;
}p[maxn];
int mp[maxn][maxn], turn_num, vis[maxn][maxn][4], nn, mm;
ll x[maxn], y[maxn];
int dir[4][2] = {1, 0, 0, -1, -1, 0, 0, 1};
//下，左，上，右
bool dfs(int xx, int yy, int turn){ //将要朝向turn
    if(vis[xx][yy][turn]) return false;
    vis[xx][yy][turn] = 1;
    int xxx = xx + dir[turn][0], yyy = yy + dir[turn][1];
    if(xxx < 0 || xxx > nn - 1 || yyy < 0 || yyy > mm - 1)
        return true;
    while(mp[xxx][yyy] != 1){
        if(vis[xxx][yyy][turn]) return false;
        vis[xxx][yyy][turn] = 1;
        xxx += dir[turn][0];
        yyy += dir[turn][1];
        if(xxx < 0 || xxx > nn - 1 || yyy < 0 || yyy > mm - 1)
            return true;
    }

    turn_num++;
    return dfs(xxx - dir[turn][0], yyy - dir[turn][1], (turn + 1) % 4);
}

int main(){
    int n, sx, sy;
    while(scanf("%d", &n) != EOF){
        turn_num = 0;
        memset(mp, 0, sizeof(mp));
        memset(vis, 0, sizeof(vis));
        for(int i = 0; i < n; i++){
            scanf("%lld%lld", &p[i].x, &p[i].y);
            x[i] = p[i].x;
            y[i] = p[i].y;
        }
        x[n] = 0, y[n] = 0, p[n].x = 0, p[n].y = 0;
        n++;
        sort(x, x + n);
        sort(y, y + n);
        int num1 = unique(x, x + n) - x;
        int num2 = unique(y, y + n) - y;
        for(int i = 0; i < n; i++){
            int X, Y;
            X = lower_bound(x, x + num1, p[i].x) - x;
            Y = lower_bound(y, y + num2, p[i].y) - y;
            if(p[i].x == 0 && p[i].y == 0){
                sx = X, sy = Y;
            }
            else{
                mp[X][Y] = 1;
            }
        }
        nn = num1, mm = num2;
        bool flag = dfs(sx, sy, 0);
        if(flag) printf("%d\n", turn_num);
        else printf("-1\n");
    }
    return 0;
}
/*
4
1 0
0 1
0 -1
-1 0
*/