BFS

BFS -> 僵尸感染 -> 队列

每个点状态的数量 ，每个状态访问一次，想办法将状态转换为int（类似hash）

int visted = [x][y][状态数量];
Queue queue;

void bfs(起点){
if起点符合退出条件，return

  起点入队  while(队列不为空){      当前临时点 = queue出队      for（各个方向遍历）{          根据当前临时点得到下一个点的位置         if(下一个点的visted没有访问过&&其他条件成立)             下一个点入队             更新下一个的信息(比如到达step)             如果符合退出条件，return             else 继续         }      }  }

}

POJ 1324

#include <cstdio>#include <iostream>using namespace std;#define SIZE 20#define LENGTH 8#define SNAKE_STATES 16384 //状态数量，以蛇头为起点，向后的每个点都有4中状态（取整），就提取了特征值int snake[LENGTH][2];int stone[SIZE * SIZE][2];int visited[SIZE + 1][SIZE + 1][SNAKE_STATES];typedef struct Element{    int x;    int y;    int state;    int step;} Element;Element Queue[SIZE * SIZE * SNAKE_STATES];int head, tail;int direct[3][3] = {    0, 0, 0,    3, 0, 1,    0, 2, 0,};int dx[4] = {-1, 0, 1, 0};int dy[4] = {0, 1, 0, -1};int getState(int Len){    int state = 0;    for(int i = Len - 1; i > 0; i--){        int x = snake[i][0] - snake[i - 1][0];        int y = snake[i][1] - snake[i - 1][1];        x+=1;        y+=1;        int D = direct[x][y];        state = state * 4 + D;    }    return state;}int isStone(int x, int y, int K){    for(int i = 0; i < K; i++){        if(x == stone[i][0] && y == stone[i][1])            return 1;    }    return 0;}int isSnake(int x, int y, int state, int L, int nx, int ny){    snake[0][0] = x;    snake[0][1] = y;    for(int i = 1; i < L; i++){        int D = state % 4;        state = state / 4;        snake[i][0] = snake[i - 1][0] + dx[D];        snake[i][1] = snake[i - 1][1] + dy[D];    }    for(int i = 0; i < L; i++){        if(nx == snake[i][0] && ny == snake[i][1]){            return 1;        }    }    return 0;}int getMinStep(int M, int N, int L, int K){    Queue[tail].x = snake[0][0];    Queue[tail].y = snake[0][1];    Queue[tail].state = getState(L);    Queue[tail].step = 0;    visited[Queue[tail].x][Queue[tail].y][Queue[tail].state] = 1;    if(Queue[tail].x == 1 && Queue[tail].y == 1) return 0;    tail++;    while(head < tail){        Element curState = Queue[head++];        //cout << "out " <<  curState.x << " " << curState.y << " " << curState.step << " " << curState.state << endl;        for(int i = 0; i < 4; i++){            int nx = curState.x + dx[i];            int ny = curState.y + dy[i];            int nd;            if(i >= 2) nd = i - 2;            else nd = i + 2;            if(nx >= 1 && nx <= M && ny >= 1 && ny <= N){                int mask = (1 << (2 * L - 2)) - 1;                int nState = ((curState.state * 4) & mask) + nd;                if(!visited[nx][ny][nState] && !isStone(nx, ny, K) && !isSnake(curState.x, curState.y, curState.state, L, nx, ny)){                    Queue[tail].x = nx;                    Queue[tail].y = ny;                    Queue[tail].state = nState;                    Queue[tail].step = curState.step + 1;                    //cout << "in " <<  nx << " " << ny << " " << Queue[tail].step << " " << nState << endl;                    if(nx == 1 && ny == 1) return Queue[tail].step;                    tail++;                    visited[nx][ny][nState] = 1;                }            }        }    }    return -1;}int main(){    //freopen("input.txt", "r", stdin);    int nC = 1;    while(1){        int M, N, L;        cin >> M >> N >> L;        if(M + N + L == 0) break;        for(int i = 0; i < L; i++){            cin >> snake[i][0] >> snake[i][1];        }        int K;        cin >> K;        for(int i = 0; i < K; i++){            cin >> stone[i][0] >> stone[i][1];        }        head = 0;        tail = 0;        for(int i = 0; i <= M; i++){            for(int j = 0; j <= N; j++){                for(int k = 0; k < SNAKE_STATES; k++){                    visited[i][j][k] = 0;                }            }        }        int minStep = getMinStep(M, N, L, K);        cout << "Case " << nC <<": "<< minStep << endl;        nC++;    }    return 0;}