HDU 1010 Tempter of the Bone

http://acm.hdu.edu.cn/showproblem.php?pid=1010大致题意：给一幅图，有起点有墙有终点，问能不能在刚好t秒的时间走到终点DFS + 多重剪枝(奇偶性剪枝)一开始果断DFS，交上去TLE了。。。用了好几重的剪枝才过。。。。所以不要小看剪枝，往往优化个成百上千倍#include<iostream>#include<cmath>using namespace std;char map[10][10];int flag, Xnum, Sx, Sy, Dx, Dy;int n, m, t;int dir[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};void DFS(int x, int y, int time){    if (x<=0 || x>n || y<=0 || y>m) return;    if (flag == 1) return;  //1.马上中断    if (x == Dx && y == Dy && time == t)    {        if(time == t)            flag = 1;        return;    }    int temp = (t - time) - abs(x - Dx) - abs(y - Dy);//2.奇偶性剪枝    if(temp<0 || temp & 1)  return;      for (int i = 0; i<4; i++)    {        int x1 = x + dir[i][0];        int y1 = y + dir[i][1];        if (map[x1][y1] != 'X')        {            map[x1][y1] = 'X';            DFS(x1, y1, time + 1);            map[x1][y1] = '.';        }    }    return;}int main(){    while (cin>>n>>m>>t)    {        if(n==0&&m==0&&t==0) break;        Xnum = 0;        for (int i = 1; i<=n; i++)        {            for (int j = 1; j<=m; j++)            {                cin>>map[i][j];                if (map[i][j] == 'S')                {                    Sx = i;                    Sy = j;                }                if (map[i][j] == 'D')                {                    Dx = i;                    Dy = j;                }                if (map[i][j] == 'X')                    Xnum ++;            }        }        flag = 0;        map[Sx][Sy] = 'X';        if (n * m - Xnum <= t)//3.提前判断t过大的情况避免再去搜，写上由500MS降到46MS        {            cout<<"NO"<<endl;            continue;        }        DFS(Sx, Sy, 0);        if (flag)            cout<<"YES"<<endl;        else             cout<<"NO"<<endl;    }    return 0;}