hdu 1026 Ignatius and the Princess I

//hdu 1026 Ignatius and the Princess I
//广搜

#include <stdio.h>
#include <string.h>
#include <queue>

using namespace std;
#define N 105
#define INF 1<<30
#define eps 1e-5
                    // x            y
const int dir[2][4] = {-1, 0, 1, 0, 0, -1, 0, 1};

struct QUE
{
    int x, y, time, step;
}que[1000000];

int line, row, head, cnt, tail;
int pre[100000];
char map[N][N];

bool no_overmap(int x, int y)
{
    if(x >= 0 && x < line && y >= 0 && y < row)
        return true;
    return false;
}

void add(int pri, int x, int y, int step, int time)
{
    pre[++head] = pri;
    que[head].x = x;
    que[head].y = y;
    que[head].step = step;
    que[head].time = time;
}

bool bfs()
{
    head = 0;
    tail = 0;
    que[++head].x = que[head].y = que[head].step = que[head].time = 0;
    map[0][0] = 'X';
    while(tail < head)
    {
        tail++;     //出队
        int x = que[tail].x, y = que[tail].y,
            step = que[tail].step;

        if(que[tail].time > 0)  //若有怪物，要杀完才能过去，这一步就必须停着
        {                   //将杀怪时间减一秒，所需时间加一秒，放入队列
            add(tail, x, y, step + 1, que[tail].time - 1);
            continue;
        }
        else
        {           //刚从队列里出来的状态若到出口,则放回
            if(x == line - 1 && y == row - 1)
                return true;
            for(int i = 0; i < 4; ++i)
            {
                int nx = x + dir[0][i], ny = y + dir[1][i];
                if(no_overmap(nx, ny) && map[nx][ny] != 'X')
                {
                    int time;
                    if(map[nx][ny] >= '0' && map[nx][ny] <= '9')
                        time = map[nx][ny] - '0';
                    else
                        time = 0;
                    add(tail, nx, ny, step + 1, time);

                    map[nx][ny] = 'X';
                }
            }
        }
    }
    return false;
}

void print_road(int now, int x, int y)  //回溯输出
{
    if(pre[now] != 0)
        print_road(pre[now], que[now].x, que[now].y);
    if(que[now].time > 0)
        printf("%ds:FIGHT AT (%d,%d)\n", cnt++, que[now].x, que[now].y);
    else
        printf("%ds:(%d,%d)->(%d,%d)\n", cnt++, que[now].x, que[now].y, x, y);
}

int main()
{
    while(scanf("%d%d", &line, &row) != EOF)
    {
        for(int i = 0; i < line; ++i)
        {
            getchar();
            for(int j = 0; j < row; ++j)
                map[i][j] = getchar();
        }
        if(bfs())
        {
            printf("It takes %d seconds to reach the target position, let me show you the way.\n", que[tail].step);
            cnt = 1;
            print_road(pre[tail], line - 1, row - 1);
        }
        else
            printf("God please help our poor hero.\n");
        puts("FINISH");
    }
    return 0;
}

福建农林大学OJ也有一题和这题差不多

http://acm.fafu.edu.cn/problem.php?id=1190，就是没有Special Judge

要按一定顺序才可以，不能用STL的priority_queue  ，而杭电的可以，因为若把节点放进队列

队列会重新排序，就不会按一定的方向出队了，我这也不是用优先队列做的，可在fafu 一直wa