/*
總時間限制: 1000ms 內存限制: 65536kB
描述 作者:1300012964
Michael喜歡滑雪百這並不奇怪, 因爲滑雪的確很刺激。可是爲了獲得速度,滑的區域必須向下傾斜,而且當你滑到坡底,你不得不再次走上坡或者等待升降機來載你。Michael想知道載一個區域中最長的滑坡。區域由一個二維數組給出。數組的每個數字代表點的高度。下面是一個例子
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
一個人可以從某個點滑向上下左右相鄰四個點之一,當且僅當高度減小。在上面的例子中,一條可滑行的滑坡爲24-17-16-1。當然25-24-23-...-3-2-1更長。事實上,這是最長的一條。
輸入
輸入的第一行表示區域的行數R和列數C(1 <= R,C <= 100)。下面是R行,每行有C個整數,代表高度h,0<=h<=10000。
輸出
輸出最長區域的長度。
樣例輸入
5 5
1 2 3 4 5
16 17 18 19 6
15 24 25 20 7
14 23 22 21 8
13 12 11 10 9
樣例輸出
25
*/
#include<iostream>
#include<cstdio>
using namespace std;
int **snow_Array;
int R,C,iMax = -1;
int **path,**trail;
int dx[] = {-1,0,0,1};
int dy[] = {0,-1,1,0};
int find_path(int x,int y)
{
int temp = 1,t;
trail[x][y] = 1;
for(int i = 0; i < 4; ++i)
{
if(x+dx[i] < 0 || y+dy[i] < 0 || x+dx[i] >= R || y+dy[i] >= C)
continue;
if(snow_Array[x][y] <= snow_Array[x+dx[i]][y+dy[i]])
continue;
if(trail[x+dx[i]][y+dy[i]] == 0)
{
<span style="white-space:pre"> </span>t = find_path(x+dx[i],y+dy[i]) + 1;
trail[x+dx[i]][y+dy[i]] = 1;
}
else
t = path[x+dx[i]][y+dy[i]] + 1;
if(temp < t)
temp = t;
}
path[x][y] = temp;
return temp;
}
int main()
{
scanf("%d%d",&R,&C);
snow_Array = new int*[R];
path = new int* [R];
trail = new int* [R];
for(int i = 0; i < R; ++i)
{
snow_Array[i] = new int[C];
path[i] = new int[C];
memset(path[i],0,sizeof(int)*C);
trail[i] = new int[C];
memset(trail[i],0,sizeof(int)*C);
for(int j = 0; j < C; ++j)
scanf("%d",&snow_Array[i][j]);
}
for(int i = 0; i < R; ++i)
for(int j = 0; j < C; ++j)
if(trail[i][j] == 0)
find_path(i,j);
for(int i = 0; i < R; ++i)
{
for(int j = 0; j < C; ++j)
{
if(iMax < path[i][j])
iMax = path[i][j];
}
}
for(int i = 0; i < R; ++i) //
{
delete [] snow_Array[i];
delete [] path[i];
delete [] trail[i];
}
delete [] snow_Array;
delete [] path;
delete [] trail;
printf("%d\n",iMax);
system("pause");
return 0;
}
/*
思考總結:
1.事實上不需要引進trail,trail的目的是爲了記載點(x,y)是否走過,這用path是否爲0就可以標記
因爲path一旦被修改,則說明(x,y)已經走了
*/
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int **snow_Array;
int R,C,iMax = -1;
int **path;
int dx[] = {-1,0,0,1};
int dy[] = {0,-1,1,0};
int find_path(int x,int y)
{
if(path[x][y])
return path[x][y];
int temp = 1;
/*
看一同學在做四個方向比較時,不用for循環,直接比較x-1,x+1,y-1,y+1,即不用dx[]來表示,
只有四個方向,這樣很簡潔,思路清晰,eg:
if(x-1>=0 && snow_Array[x][y] > snow_Array[x-1][y])
temp = max(temp,find_path(x-1,y)+1);
if(x+1<R && snow_Array[x][y] > snow_Array[x+1][y])
temp = max(temp,find_path(x+1,y)+1);
if(y-1>=0 && snow_Array[x][y] > snow_Array[x][y-1])
temp = max(temp,find_path(x,y-1)+1);
if(y+1<C && snow_Array[x][y] > snow_Array[x][y+1])
temp = max(temp,find_path(x,y+1)+1);
————————————————————————————————————————————————————————————
而此處的for循環思路也一樣
如果點超出邊界 和 不滿足下一個點高度比這一點小,則continue
用temp與四個方向的比較
注意賦初值temp = 1
*/
for(int i = 0; i < 4; ++i)
{
if(x+dx[i] < 0 || y+dy[i] < 0 || x+dx[i] >= R || y+dy[i] >= C)
continue;
if(snow_Array[x][y] <= snow_Array[x+dx[i]][y+dy[i]])
continue;
if(temp < find_path(x+dx[i],y+dy[i]) + 1)
temp = find_path(x+dx[i],y+dy[i]) + 1;
}
path[x][y] = temp;
return temp;
}
int main()
{
scanf("%d%d",&R,&C);
snow_Array = new int*[R];
path = new int* [R];
for(int i = 0; i < R; ++i)
{
snow_Array[i] = new int[C];
path[i] = new int[C];
memset(path[i],0,sizeof(int)*C);
for(int j = 0; j < C; ++j)
scanf("%d",&snow_Array[i][j]);
}
/*
這一步其實也不需要這麼麻煩,因爲find_path本來就是計算(x,y)可到達的長度的函數
就不需要在這裏再走,這就修改find_path函數,如果(x,y)已走,返回path[x][y];
for(int i = 0; i < R; ++i)
for(int j = 0; j < C; ++j)
if(path[i][j] == 0)
find_path(i,j);
*/
for(int i = 0; i < R; ++i)
{
for(int j = 0; j < C; ++j)
{
if(iMax < find_path(i,j))
iMax = find_path(i,j);
}
}
for(int i = 0; i < R; ++i) //釋放內存
{
delete [] snow_Array[i];
delete [] path[i];
}
delete [] snow_Array;
delete [] path;
printf("%d\n",iMax);
system("pause");
return 0;
}