hdu1071Nightmare(BFS+优先队列)

题目链接： http://acm.hdu.edu.cn/showproblem.php?pid=1072

这个题用普通的队列也可以，关键的地方是 对于炸弹重置装置只能访问一次，虽然题中允许访问多次。

我的代码：

#include <cstdio>
#include <queue>
#include <cstring>

using namespace std;

int n,m;
int a[50][50];
int stx,sty,endx,endy;
int flag;
struct node{
    int x,y;
    int time;
    int step;
};
node ans;
struct cmp{
    bool operator()(node x,node y)  // x.step比y.step优先级小 返回true
    {
        return x.step > y.step;
    }
};
int dx[] = {-1,0,1,0};
int dy[] = {0,1,0,-1};


void bfs()

{
    priority_queue<node,vector<node>,cmp> q;
    node f ;
    f.x = stx;
    f.y = sty;
    f.time = 6;
    f.step = 0;
    q.push(f);
    while(!q.empty())
    {
        node h = q.top();
        q.pop();
        for(int i = 0;i < 4;++i)
        {
            node t;
            int tx = h.x + dx[i];
            int ty = h.y + dy[i];
            if(tx>=0&&tx<n&&ty>=0&&ty<m&&a[tx][ty])
            {
                if(a[tx][ty] == 4 && h.time - 1)
                {
                    t.x = tx;
                    t.y = ty;
                    t.time = 6;
                    t.step = h.step + 1;
                    a[tx][ty] = 0;
                    q.push(t);
                }
                else if(a[tx][ty] == 1 || a[tx][ty] == 3)
                {
                    t.x = tx;
                    t.y = ty;
                    t.time = h.time - 1;
                    t.step = h.step + 1;
                    if(tx == endx && ty == endy && t.time)
                    {
                        flag = 1;
                        ans = t;
                        return ;
                    }
                    if(t.time){
                        q.push(t);
                    }
                }
            }
        }
    }
}
int main()

{
    int _;
    scanf("%d",&_);
    while(_--)
    {
        scanf("%d%d",&n,&m);
        for(int i = 0;i < n;++i)
            for(int j = 0;j < m;++j){
                scanf("%d",&a[i][j]);
                if(a[i][j] == 2)
                {
                    stx = i;
                    sty = j;
                }
                else if(a[i][j] == 3)
                {
                    endx = i;
                    endy = j;
                }
            }
        flag = 0;
        bfs();
        if(flag)
            printf("%d\n",ans.step);
        else
            printf("-1\n");
    }
    return 0;
}