50044. Caterpillar in Panic

難度：4.5/5 Used Time: 30:49

細節多。 學到的東西：

  1. \(d \in \{0, 1, 2, 3\}\)，所以可以建立

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    int dx[] = {1, -1, 0, 0}, dy[] ={0, 0, 1, -1};
    然後更新head可以使用：
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    head_x += dx[d], head_y += dy[d];

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    len = abs(sx - ex) + abs(sy - ey) + 1;
    if(sx == ex) dx = 0, dy = (sy < ey ? 1 : -1);
    else dy = 0, dx = (sx < ex ? 1 : -1);

Ver 1

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#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>

int n, m;
int sx, sy, ex, ey;
int q;
int len;
int head, tail;
int flag = true;

int map[1005][1005];
int row[1005];
int col[1005];

void print_map(){
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            printf("%d", map[i][j]);
        }
        printf("\n");
    }
}

bool is_valid(int x, int y){
    return (x >= 0 && x < n &&
            y >= 0 && y < m &&
            map[x][y] == 0);
}

void ipt(){
    scanf("%d %d", &n, &m);
    scanf("%d %d %d %d", &sx, &sy, &ex, &ey);
    scanf("%d", &q);

    if(sx == ex){
        len = abs(ey - sy) + 1;
        if(sy < ey){
            for(int i = 0; i < len; i++) row[i] = sx;
            for(int i = 0, n = sy; i < len; i++, n++) col[i] = n;
        }
        else{
            for(int i = 0; i < len; i++) row[i] = sx;
            for(int i = 0, n = sy; i < len; i++, n--) col[i] = n;
        }
    }
    else{
        len = abs(ex - sx) + 1;
        if(sx < ex){
            for(int i = 0; i < len; i++) col[i] = sy;
            for(int i = 0, n = sx; i < len; i++, n++) row[i] = n;
        }
        else{
            for(int i = 0; i < len; i++) col[i] = sy;
            for(int i = 0, n = sx; i < len; i++, n--) row[i] = n;
        }
    }

    for(int i = 0; i < n; i++) for(int j = 0; j < m; j++){
        map[i][j] = 0;
    }
    for(int i = 0; i < len; i++){
        map[row[i]][col[i]] = 1;
    }
    head = 0, tail = len - 1;

}

void move(){
    int d; scanf("%d", &d);

    int tail_x = row[tail], tail_y = col[tail];
    map[tail_x][tail_y] = 0;
    tail = tail > 0 ? tail - 1 : len - 1; // tail--;

    int head_x = row[head], head_y = col[head];
    if(d == 0) head_x += 1;
    else if(d == 1) head_x -= 1;
    else if(d == 2) head_y += 1;
    else if(d == 3) head_y -= 1;

    if(is_valid(head_x, head_y)){
        map[head_x][head_y] = 1;
        head = head > 0 ? head - 1 : len - 1;
        row[head] = head_x, col[head] = head_y;
    }
    else{
        map[tail_x][tail_y] = 1;
        flag = false;
        return;
    }
}

int main(){
    ipt();
    for(int i = 0; i < q && flag; i++) move();
    print_map();
}

Ver 2

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#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>

int n, m;
int sx, sy, ex, ey;
int q;

int len;
int head, tail;
int row[1024], col[1024];
int map[1024][1024];

bool flag = true;

void print_map(){
    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++){
            printf("%d", map[i][j]);
        }
        puts("");
    }
}
void ipt(){
    scanf("%d %d", &n, &m);
    scanf("%d %d %d %d", &sx, &sy, &ex, &ey);
    scanf("%d", &q);

    int dx, dy;
    len = abs(ex - sx) + abs(ey - sy) + 1;
    head = 0, tail = len - 1;
    if(sx == ex) dx = 0, dy = (sy < ey ? 1 : -1);
    else dy = 0, dx = (sx < ex ? 1 : -1);

    for(int i = 0; i < n; i++) for(int j = 0; j < m; j++){
        map[i][j] = 0;
    }
    for(int i = 0, now_x = sx, now_y = sy; i < len; i++, now_x += dx, now_y += dy){
        map[now_x][now_y] = 1;
        row[i] = now_x, col[i] = now_y;
    }
}

bool is_valid(int x, int y){
    return (x >= 0 && x < n &&
            y >= 0 && y < m &&
            map[x][y] == 0);
}

void move(){
    int d; scanf("%d", &d);
    int dx[] = {1, -1, 0, 0}, dy[] = {0, 0, 1, -1};

    int tail_x = row[tail], tail_y = col[tail];
    map[tail_x][tail_y] = 0;
    tail = tail > 0 ? tail - 1 : len - 1;

    int head_x = row[head], head_y = col[head];
    head_x += dx[d], head_y += dy[d];
    if(is_valid(head_x, head_y)){
        map[head_x][head_y] = 1;
        head = head > 0 ? head - 1 : len - 1;
        row[head] = head_x, col[head] = head_y;
    }
    else{
        map[tail_x][tail_y] = 1;
        flag = false;
    }
}

int main(){
    ipt();
    for(int i = 0; i < q && flag; i++){
        move();
    }
    print_map();
    return 0;
}

Coding > Solutions > Judge Girl > 2016
#2d-array
50044. Caterpillar in Panic
https://aaronlin1229.github.io/judgegirl_50044/
Author
Akizumi
Posted on
July 17, 2023
Licensed under