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）
——>>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：
[cpp] #include
#include
#include
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;
}
#include
#include
#include
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：
[cpp] #include
#include
#include
#include
using namespace std;
const int maxn = 7