2022牛客多校第三场 J题 Journey

题目链接

Journey

题目大意

给了我们一堆十字路口，问我们从一条走到另一条路消耗的最少的体力，每次向右转可以不消耗体力，其余方向均为消耗$1$点体力。

题解

理解清楚题意后，直接双端队列广搜即可。

代码

#include<iostream>
#include<cstring>
#include<queue>
#include<deque>
#include<map>
using namespace std;
const int maxn = 500010;
#define x first
#define y second
typedef pair<int,int> PII;
int n;
int sx, sy, ex, ey;
int w[maxn][4];
map<PII, int> dist;
typedef struct node
{
    
    int x, y, d;
}Node;
int get_direction(int a, int b)
{
    
    for(int i = 0; i < 4; i ++){
    
        if(w[b][i] == a) {
    
            return (i + 1) % 4;
        }
    }
    return -1;
}
int bfs()
{
    
    deque<Node> q;
    q.push_back({
    sx, sy, 0});
    while(q.size())
    {
    
        auto t = q.front();
        q.pop_front();
        if(t.x == ex && t.y == ey) return t.d;
        int d = get_direction(t.x, t.y);
        for(int i = 0; i < 4; i ++){
    
            int j = w[t.y][i];
            if(j == 0) continue;
            int v = 1;
            if(i == d) v = 0;
            if(!dist.count({
    t.y, j}) || dist[{
    t.y, j}] > t.d + v) {
    
                dist[{
    t.y, j}] = t.d + v;
                if(v == 0) q.push_front({
    t.y, j, dist[{
    t.y, j}]});
                else q.push_back({
    t.y, j, dist[{
    t.y, j}]});
            }
        }
    }
    return -1;
}
int main()
{
    
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++){
    
        for(int j = 0; j < 4; j ++){
    
            scanf("%d", &w[i][j]);
        }
    }
    scanf("%d %d %d %d", &sx, &sy, &ex, &ey);
    printf("%d\n", bfs());
    return 0;
}

把map换成unordered_map会快超多！

#include<iostream>
#include<cstring>
#include<queue>
#include<deque>
#include<map>
#include<unordered_map>
using namespace std;
const int maxn = 500010;
#define x first
#define y second
typedef pair<int,int> PII;
int n;
int sx, sy, ex, ey;
int w[maxn][4];
unordered_map<int, int> dist[maxn];
typedef struct node
{
    
    int x, y, d;
}Node;
int get_direction(int a, int b)
{
    
    for(int i = 0; i < 4; i ++){
    
        if(w[b][i] == a) {
    
            return (i + 1) % 4;
        }
    }
    return -1;
}
int bfs()
{
    
    deque<Node> q;
    q.push_back({
    sx, sy, 0});
    while(q.size())
    {
    
        auto t = q.front();
        q.pop_front();
        if(t.x == ex && t.y == ey) return t.d;
        int d = get_direction(t.x, t.y);
        for(int i = 0; i < 4; i ++){
    
            int j = w[t.y][i];
            if(j == 0) continue;
            int v = 1;
            if(i == d) v = 0;
            if(!dist[t.y].count(j) || dist[t.y][j]> t.d + v) {
    
                dist[t.y][j] = t.d + v;
                if(v == 0) q.push_front({
    t.y, j, dist[t.y][j]});
                else q.push_back({
    t.y, j, dist[t.y][j]});
            }
        }
    }
    return -1;
}
int main()
{
    
    scanf("%d", &n);
    for(int i = 1; i <= n; i ++){
    
        for(int j = 0; j < 4; j ++){
    
            scanf("%d", &w[i][j]);
        }
    }
    scanf("%d %d %d %d", &sx, &sy, &ex, &ey);
    printf("%d\n", bfs());
    return 0;
}