50063. Friend Distance

難度：4.3/5

Second Try: 4/5 Used time: 32:41 I though the problem require us to add up all the friend pairs' distance, but it was the max distance for a pair, I really need to look at the problem carefully.

Version 1 code (TLE > 2s)

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#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
 
int n, m;
int arr[16][16] = {0};
int min_dist = INT_MAX;
int min_pos[16];
 
static inline int my_abs(int n){
    if(n < 0) return -n;
    else return n;
}
static inline int max(int a, int b){
    if(a < b) return b;
    return a;
}
static inline int min(int a, int b){
    if(a < b) return a;
    return b;
}
 
void dfs(int pos[], int now_idx, int now_max_dist){
    if(now_max_dist >= min_dist) return;
 
    if(now_idx == n){
        min_dist = now_max_dist;
        for(int i = 0; i < n; i++){
            min_pos[i] = pos[i];
        }
    }
    else{
        for(int i = 0; i < n; i++){
            if(pos[i] == -1){
                pos[i] = now_idx;
                int temp_max_dist = now_max_dist;
                for(int j = 0; j < n; j++){
                    if(arr[i][j] == 1 && pos[j] != -1){
                        temp_max_dist = max(temp_max_dist, my_abs(now_idx - pos[j]));
                    }
                }
                dfs(pos, now_idx + 1, temp_max_dist);
                pos[i] = -1;
            }
        }
    }
}
 
int main(){
    scanf("%d %d", &n, &m); getchar();
    for(int i = 0; i < m; i++){
        char a, b;
        a = getchar(); getchar();
        b = getchar(); getchar();
        arr[a - 'A'][b - 'A'] = 1;
        arr[b - 'A'][a - 'A'] = 1;
    }
 
    int pos[16]; for(int i = 0; i < n; i++) pos[i] = -1;
    dfs(pos, 0, 0);
 
    printf("%d\n", min_dist);
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(min_pos[j] == i){
                if(i == 0) printf("%c", j + 'A');
                else printf(" %c", j + 'A');
            }
        }
    }
    printf("\n");
}

Version 2 code (AC 800ms)

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#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#define max(a, b) (((a) > (b)) ? (a) : (b))

int n, m;
int arr[16][16] = {0};
int min_dist = INT_MAX;
int min_pos[16];

void dfs(int now[], int now_idx, int used[], int now_max_dist){
    if(now_max_dist >= min_dist) return;
    if(now_idx == n){
        min_dist = now_max_dist;
        for(int i = 0; i < n; i++) min_pos[i] = now[i];
    }
    else{
        for(int i = 0; i < n; i++){
            if(used[i] == 0){
                used[i] = 1, now[now_idx] = i;

                int temp_max_dist = 0;
                for(int j = 0; j < now_idx; j++){
                    if(arr[i][now[j]]){
                        temp_max_dist = now_idx - j;
                        break;
                    }
                }
                temp_max_dist = max(temp_max_dist, now_max_dist);
                dfs(now, now_idx + 1, used, temp_max_dist);
                used[i] = 0, now[now_idx] = 0;
            }
        }
    }
}

int main(){
    scanf("%d %d", &n, &m); getchar();
    for(int i = 0; i < m; i++){
        char a, b;
        a = getchar(); getchar();
        b = getchar(); getchar();
        arr[a - 'A'][b - 'A'] = 1;
        arr[b - 'A'][a - 'A'] = 1;
    }

    int now[16] = {0};
    int used[16] = {0};
    dfs(now, 0, used, 0);

    printf("%d\n", min_dist);
    printf("%c", min_pos[0] + 'A');
    for(int i = 1; i < n; i++) printf(" %c", min_pos[i] + 'A');
    printf("\n");
    return 0;

}

Coding > Solutions > Judge Girl > 2016
#recursion
50063. Friend Distance
https://aaronlin1229.github.io/judgegirl_50063/
Author
Akizumi
Posted on
July 17, 2023
Licensed under