hdu - 1010 - Tempter of the Bone

题意：一个N*M的地图，走过的点不能再走，X为墙不可走，能否从点S到点D恰好用时T。（1 < N, M < 7; 0 < T < 50）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1010

——>>有点邯郸学步的感觉，寒假时用cin输入数据，轻松AC，后来慢慢改成了用scanf来输入，今天，这道坑爹的题目交了n次也是WA，代替了n种方式后发现，是输入的问题，猜想是输入的地图有空格……

超时优化的思想：

1、可走格子数少于T时，一定不ok；

2、从格子A到格子B，无论怎么走，步数的奇偶性相同。

用时很少的写法：

#include <cstdio>
#include <cmath>
#include <iostream>

using namespace std;

char MAP[8][8];
int N, M, T, s_x, s_y, d_x, d_y;
int dx[] = {-1, 0, 1,  0};
int dy[] = { 0, 1, 0, -1};

bool ok;

void dfs(int x, int y, int cur)
{
    if(ok) return;
    int temp = T - cur - abs(d_x-x) - abs(d_y-y);
    if((temp < 0) || (temp % 2 == 1)) return;
    for(int i = 0; i < 4; i++)
    {
        int new_x = x + dx[i];
        int new_y = y + dy[i];
        if(new_x >= 0 && new_x < N && new_y >= 0 && new_y < M && MAP[new_x][new_y] != 'X')
        {
            if(new_x == d_x && new_y == d_y && cur+1 == T) ok = 1;
            else
            {
                MAP[new_x][new_y] = 'X';
                dfs(new_x, new_y, cur+1);
                MAP[new_x][new_y] = '.';
            }
        }
    }
}
int main()
{
    int i, j, sum;
    while(scanf("%d%d%d", &N, &M, &T) == 3)
    {
        if(N == 0 && M == 0 && T == 0) return 0;
        sum = 1;
        for(i = 0; i < N; i++)
        {
            //getchar();
            for(j = 0; j < M; j++)
            {
                //MAP[i][j] = getchar();
                //scanf("%c", &MAP[i][j]);
                cin>>MAP[i][j];
                if(MAP[i][j] == 'S')
                {
                    s_x = i;
                    s_y = j;
                }
                else if(MAP[i][j] == 'D')
                {
                    d_x = i;
                    d_y = j;

                }
                else if(MAP[i][j] == 'X') sum++;
            }

        }
        sum = N*M - sum;
        ok = 0;
        MAP[s_x][s_y] = 'X';
        if(T <= sum)
            dfs(s_x, s_y, 0);
        if(ok) printf("YES\n");
        else printf("NO\n");
    }
    return 0;
}

vis[][]数组写法：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>

using namespace std;

const int maxn = 7 + 10;
int dx[] = {-1, 1,  0, 0};
int dy[] = { 0, 0, -1, 1};
char MAP[maxn][maxn];
bool ok, vis[maxn][maxn];
int N, M, T, Dx, Dy;

void dfs(int x, int y, int cur)
{
    if(ok) return;
    if(cur == T)
    {
        if(x == Dx && y == Dy)
        {
            ok = 1;
            return;
        }
        else
            return;
    }
    int temp = T - cur - (Dx - x) - (Dy - y);
    if(temp < 0 || (temp&1)) return;
    for(int i = 0; i < 4; i++)
    {
        int newx = x + dx[i];
        int newy = y + dy[i];
        if(newx >= 0 && newx < N && newy >= 0 && newy < M && MAP[newx][newy] != 'X' && !vis[newx][newy])
        {
            vis[newx][newy] = 1;
            dfs(newx, newy, cur+1);
            vis[newx][newy] = 0;
        }
    }
}

int main()
{
    int Sx, Sy, i, j;
    while(scanf("%d%d%d", &N, &M, &T) == 3)
    {
        if(!N && !M && !T) return 0;
        for(i = 0; i < N; i++)
        {
            for(j = 0; j < M; j++)
            {
                cin>>MAP[i][j];
                if(MAP[i][j] == 'S')
                {
                    Sx = i;
                    Sy = j;
                }
                else if(MAP[i][j] == 'D')
                {
                    Dx = i;
                    Dy = j;
                }
            }
        }
        ok = 0;
        memset(vis, 0, sizeof(vis));
        vis[Sx][Sy] = 1;
        dfs(Sx, Sy, 0);
        if(ok)
            printf("YES\n");
        else
            printf("NO\n");
    }
    return 0;
}