cf97D. Robot in Basement(模拟 bitset)

题意

题目链接

Sol

接下来我的实现方式和论文里不太一样

然后用bitset优化，上下走分别对应着右移/左移m位，左右走对应着右移/左移1位

我们可以直接预处理出能走的格子和不能走的格子，每次走的时候先全都走过去，再把撞到墙上的补回来即可

#include<bits/stdc++.h>
#define u32 unsigned int
using namespace std;
const int MAXN = 1e5 + 10, SS = 150 * 150 + 150;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, Len, id;
bitset<SS> now, B, can;
char s[151], str[MAXN];
int main() {
    N = read(); M = read(); Len = read();
    for(int i = 1; i <= N; i++) {
        scanf("%s", s + 1);
        for(int j = 1; j <= M; j++) {
            if(s[j] == '#') B[i * M + j] = 1;
            else can[i * M + j] = now[i * M + j] = 1;
            if(s[j] == 'E') id = i * M + j;
        }
    }
    scanf("%s", str + 1);
    if(now.count() == 1 && now[id] == 1) return puts("0"), 0;
    for(int i = 1; i <= Len; i++) {
        if(str[i] == 'U') {
            now = ((now >> M) & can) | (now & (B << M));
        } else if(str[i] == 'D') {
            now = ((now << M) & can) | (now & (B >> M));
        } else if(str[i] == 'L') {
            now = ((now >> 1) & can) | (now & (B << 1));
        } else if(str[i] == 'R') {
            now = ((now << 1) & can) | (now & (B >> 1));
        }
        if(now.count() == 1 && now[id] == 1) {
            printf("%d\n", i);
            return 0;
        }
    }
    puts("-1");
    return 0;
}