• CodeForces 676D代码 哪里有问题呢?


    题目:

    http://codeforces.com/problemset/problem/676/D

    code:

    #include <stdio.h>
    #define MAXN 1001
    #define LEFT 0
    #define TOP 1
    #define RIGHT 2
    #define BOTTOM 3
    struct block{
        //left,top,right,bottom
        int doors[4];
        int doors_num;
        int rotats;
        int minutes;
    };
    struct block labyrinth[MAXN][MAXN];
    char visited[MAXN][MAXN];
    
    struct node{
        int x;
        int y;
    };
    int top;
    struct node path[MAXN*MAXN];
    int N,M;
    int xt,yt,xm,ym;
    int min_minutes = 0x0FFFFFFF;
    void init_doors(int x,int y,char c)
    {
        int i;
        for (i=0;i<4;i++)
        {
            labyrinth[x][y].doors[i] = 0;
        }
        labyrinth[x][y].doors_num =0;
        labyrinth[x][y].rotats = 0;
        labyrinth[x][y].minutes = 0x0FFFFFFF;
        switch(c)
        {
        case '+':
            for (i=0;i<4;i++)
            {
                labyrinth[x][y].doors[i] = 1;
            }
            labyrinth[x][y].doors_num = 4;
            labyrinth[x][y].rotats = 0;
            break;
        case '-':
            labyrinth[x][y].doors[LEFT] = 1;
            labyrinth[x][y].doors[RIGHT] =1;
            labyrinth[x][y].doors_num = 2;
            break;
        case '|':
            labyrinth[x][y].doors[TOP] = 1;
            labyrinth[x][y].doors[BOTTOM] =1;
            labyrinth[x][y].doors_num = 2;
            break;
        case '^':
            labyrinth[x][y].doors[TOP] = 1;
            labyrinth[x][y].doors_num = 1;
            break;
        case '>':
            labyrinth[x][y].doors[RIGHT] = 1;
            labyrinth[x][y].doors_num = 1;
            break;
        case '<':
            labyrinth[x][y].doors[LEFT] = 1;
            labyrinth[x][y].doors_num = 1;
            break;
        case 'V':
            labyrinth[x][y].doors[BOTTOM] = 1;
            labyrinth[x][y].doors_num = 1;
            labyrinth[x][y].rotats = 0;
            break;
        case 'L':
            labyrinth[x][y].doors[TOP] = 1;
            labyrinth[x][y].doors[BOTTOM] =1;
            labyrinth[x][y].doors[RIGHT] = 1;
            labyrinth[x][y].doors_num =3;
            labyrinth[x][y].rotats = 0;
            break;
        case 'R':
            labyrinth[x][y].doors[TOP] = 1;
            labyrinth[x][y].doors[BOTTOM] =1;
            labyrinth[x][y].doors[LEFT] = 1;
            labyrinth[x][y].doors_num = 3;
            labyrinth[x][y].rotats = 0;
            break;
        case 'U':
            labyrinth[x][y].doors[RIGHT] = 1;
            labyrinth[x][y].doors[BOTTOM] =1;
            labyrinth[x][y].doors[LEFT] = 1;
            labyrinth[x][y].doors_num = 3;
            break;
        case 'D':
            labyrinth[x][y].doors[TOP] = 1;
            labyrinth[x][y].doors[RIGHT] =1;
            labyrinth[x][y].doors[LEFT] = 1;
            labyrinth[x][y].doors_num = 3;
            break;
        }
    }
    void push(int i,int j)
    {
        top++;
        path[top].x = i;
        path[top].y = j;
        
    }
    struct node pop()
    {
        top--;
        return path[top +1];
    }
    int get_door(int x,int y,int direction,int roates)
    {
        int t = roates % 4;
        int index = direction - t;
        if (index < 0)
        {
            index += 4;
        }
    
        return labyrinth[x][y].doors[index];
    
    }
    void process_state(int x,int y,int cur_rotates,int cur_time)
    {
        int dir_y[4] = {-1,0,1,0};
        int dir_x[4] = { 0,-1,0,1};
        int i,nx,ny;
        for (i=0;i<4;i++)
        {
            nx = x+dir_x[i];
            ny = y+dir_y[i];
            if (nx<1 || nx >N)
            {
                continue;
            }
            if (ny<1 || ny>M)
            {
                continue;
            }
            if (get_door(x,y,i,cur_rotates)&& get_door(nx,ny,(i+2)%4,cur_rotates))
            {
                if (!visited[nx][ny])
                {
                    visited[nx][ny] =1;
                    labyrinth[nx][ny].minutes = cur_time+1;
                    labyrinth[nx][ny].rotats = cur_rotates;
                    if (nx == xm && ny == ym)
                    {
                        if (min_minutes > cur_time+1)
                        {
                            min_minutes = cur_time+1;
                        }
                    }else
                    {
    
                        push(nx,ny);
                    }
    
                }else
                {
                    if (labyrinth[nx][ny].minutes >cur_time+1 )
                    {
                        labyrinth[nx][ny].minutes = cur_time+1;
                        labyrinth[nx][ny].rotats = cur_rotates;
                        push(nx,ny);
                    }
                }
            }
    
        }
    }
    int bfs()
    {
        int x,y,cur_rotates,i,nx,ny,t,cur_time;
        
        struct node cur;
        top = -1;
        push(xt,yt);
        labyrinth[xt][yt].minutes = 0;
        visited[xt][yt] =1;
        
        while(top!= -1)
        {
            cur = pop();
    
            //left
            x = cur.x;
            y = cur.y;
            cur_rotates = labyrinth[x][y].rotats;
            cur_time = labyrinth[x][y].minutes;
            t = 0;
            while(t < 4)
            {
                if (cur_time + t < min_minutes)
                {
                    process_state(x,y,cur_rotates+t,cur_time+t);
                }
                t++;
                
            }
        }
        return labyrinth[xm][ym].minutes;
            
    }
    int main()
    {
        int i,j;
        char str[MAXN];
    scanf(
    "%d %d",&N,&M); for (i=1;i<=N;i++) { scanf("%s ",str); for (j=1;j<=M;j++) { init_doors(i,j,str[j-1]); visited[i][j] = 0; } } scanf("%d %d",&xt,&yt); scanf("%d %d",&xm,&ym); //check same logcation if (xt == xm && yt == ym) { printf("0"); return 0; } //dfs(xt,yt,0); bfs(); if(min_minutes == 0x0FFFFFFF) printf("-1"); else printf("%d",min_minutes); return 0; }
  • 相关阅读:
    SwiftUI使用URLSession发送https请求免证书验证
    IOS 时间格式 格式化说明
    句柄,文件描述符的理解
    Docker Swarm(8)
    Docker 自己的理解(9)
    eclipse导入idea项目
    写写最近的感悟为什么好久没更
    linux 命令行实现CPU 100%占用
    2.如何正确理解古典概率中的条件概率《zobol的考研概率论教程》
    1.为什么要从古典概率入门概率学《zobol的考研概率论教程》
  • 原文地址:https://www.cnblogs.com/xiaoyu-10201/p/5535733.html
Copyright © 2020-2023  润新知