HDU 5040 Instrusive(2014 ACM/ICPC Asia Regional Beijing Online )

题目分析

这道题的主要关键就是怎样定义一个能判断重复的状态，很明显，方向每4s一个循环，mod4即可，然后根据所在的x和y坐标就可以定义出状态state[4][505][505]，关于在不在cardbox中无所谓，因为这个不需要时间去切换，那么很明显我们对于需要的时候就用cardbox，不需要的时候直接不用就可以了。

#include <queue>#include <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const int maxn = 505;const int loc[4][2] = {{1,0},{-1,0},{0,-1},{0,1}};char maze[maxn][maxn];  //迷宫int vis[4][maxn][maxn]; //判断摄像头的位置能不能走，只可能有4种状态int state[4][maxn][maxn];  //标记状态int T, N, sx, sy;struct Node{  //节点，往priority_queue里面放，重载运算符    int x, y, dir, t;    Node() {}    Node(int a,int b,int c,int d):x(a), y(b), dir(c), t(d){}    bool operator < (const Node temp)const{        return t > temp.t;    }};void change(){  //根据时间改变即可    for(int i = 0; i < N; i++)        for(int j = 0; j < N; j++)            if(maze[i][j] == 'N') maze[i][j] = 'E';            else if(maze[i][j] == 'S') maze[i][j] = 'W';            else if(maze[i][j] == 'W') maze[i][j] = 'N';            else if(maze[i][j] == 'E') maze[i][j] = 'S';            else if(maze[i][j] == 'M') sx = i, sy = j;}void init(){  //初始化    memset(vis, 0, sizeof(vis));    memset(state, 0, sizeof(state));    for(int k = 0; k < 4; k++){        for(int i = 0; i < N; i++){            for(int j = 0; j < N; j++){                if(maze[i][j] == 'N'){                    if(i >= 1) vis[k][i][j] = 1, vis[k][i-1][j] = 1;                    else vis[k][i][j] = 1;                }                else if(maze[i][j] == 'S'){                    vis[k][i][j] = 1;                    vis[k][i+1][j] = 1;                }                else if(maze[i][j] == 'W'){                    if(j >= 1) vis[k][i][j-1] = 1, vis[k][i][j] = 1;                    else vis[k][i][j] = 1;                }                else if(maze[i][j] == 'E'){                    vis[k][i][j] = 1;                    vis[k][i][j+1] = 1;                }                else if(maze[i][j] == '#') state[k][i][j] = 1;            }        }        change();    }}bool judge(int x,int y){    if(x < 0 || x >= N || y < 0 || y >= N) return false;    else return true;}int bfs(){    priority_queue <Node> pq;    Node pre, cur;    pq.push(Node(sx, sy, 0, 0));    state[0][sx][sy] = 1;    while(!pq.empty()){        Node cur = pq.top(); pq.pop();        if(maze[cur.x][cur.y] == 'T') return cur.t;        if(vis[cur.dir][cur.x][cur.y]){            for(int i = 0; i < 4; i++){                int xx = cur.x + loc[i][0];                int yy = cur.y + loc[i][1];                if(judge(xx, yy) && !state[(cur.dir+3)%4][xx][yy]){                    state[(cur.dir+3)%4][xx][yy] = 1;                    pq.push(Node(xx, yy, (cur.dir+3)%4, cur.t+3));                }            }            if(!state[(cur.dir+1)%4][cur.x][cur.y]){                state[(cur.dir+1)%4][cur.x][cur.y] = 1;                pq.push(Node(cur.x, cur.y, (cur.dir+1)%4, cur.t+1));            }        }        else{            for(int i = 0; i < 4; i++){                int xx = cur.x + loc[i][0];                int yy = cur.y + loc[i][1];                if(!judge(xx, yy)) continue;                if(!vis[cur.dir][xx][yy]){                    if(!state[(cur.dir+1)%4][xx][yy]){                        state[(cur.dir+1)%4][xx][yy] = 1;                        pq.push(Node(xx, yy, (cur.dir+1)%4, cur.t+1));                    }                }                else{                    if(!state[(cur.dir+3)%4][xx][yy]){                        state[(cur.dir+3)%4][xx][yy] = 1;                        pq.push(Node(xx, yy, (cur.dir+3)%4, cur.t+3));                    }                }            }            if(!state[(cur.dir+1)%4][cur.x][cur.y]){                state[(cur.dir+1)%4][cur.x][cur.y] = 1;                pq.push(Node(cur.x, cur.y, (cur.dir+1)%4, cur.t+1));            }        }    }    return -1;}int main(){    scanf("%d", &T);    for(int kase = 1; kase <= T; kase++){        scanf("%d", &N);        for(int i = 0; i < N; i++) scanf("%s", maze[i]);        init();        printf("Case #%d: %d\n", kase, bfs());    }    return 0;}