Time Limit: 1000 MS
Memory Limit: 32768 K
Description
Doctor Tang is an excellent Architect. Before awarded “the greatest Architect”, he needs to help government design a building group. The secret of mayor gives the Doctor Tang a map of platform,
2 2 1 2
2 2 1 1
3 2 1 2which means there are buildings on every unit square. Each number means the height of the building.
Then, Doctor Tang is supposed to design a draft in 3D-view like those.
<p align="center">
</p>
The first picture is viewed form the front, and the second is from the left.
We define every layer of a building follow the same format strictly. 
In order to show you the space, I turn it into a point on the picture. But you should just print the point you need to print. That’s means the point on the square shouldn’t be print. Here are some simple examples:
<p align="center">
</p>
The empty places decorate with “.”. Can you help Doctor Tang solving this problem? Just clam down, the problem is not as difficult as it seems.</td>
Input
There are multiple test cases. For each test case, the first line has two integers: N, M. Following N lines, each line has M integers h1,h2…hm, describing the height of building in this position.
(1<=N<=50, 1<=M<=100, 1<=hi<=100)
Output
For each test case, print a 3D-view viewing from the front and from the left.
Sample Input
3 4
2 2 1 2
2 2 1 1
3 2 1 2Sample Output
......+---+---+...+---+
..+---+ / /|../ /|
./ /|-+---+ |.+---+ |
+---+ |/ /| +-| | +
| | +---+ |/+---+ |/|
| |/ /| +/ /|-+ |
+---+---+ |/+---+ |/| +
| | | +-| | + |/.
| | |/ | |/| +..
+---+---+---+---+ |/...
| | | | | +....
| | | | |/.....
+---+---+---+---+......
........+---+...+---+
......./ /|../ /|
......+---+ |.+---+ |
......| +---+ | +
....+---+/ /|-+ |/|
.../ /+---+ |/|-+ |
..+---+-| | + |/| +
./ / | |/| + |/.
+---+---+---+ |/| +..
| | | | + |/...
| | | |/| +....
+---+---+---+ |/.....
| | | | +......
| | | |/.......
+---+---+---+........题目大意
输入 n m 以及nxm的位置中方块的数量。。。。
然后输出 正视图和左视图。。。。
无聊模拟啊。。。 比赛的时候建系错误 导致没AC 然后内存开始没想大多给的超大...
//还有很多可以优化的地方。。。 写的非常撮
#include
#include
int mp[55][105];
char p[1000][1000];
int n,m,height;
int prtx,prty;
void draw(int x,int y){
p[x][y+1]=p[x][y+2]=p[x][y+3]=p[x+3][y+1]=p[x+3][y+2]=p[x+3][y+3]=p[x+5][y+3]=p[x+5][y+4]=p[x+5][y+5]='-';
p[x][y]=p[x][y+4]=p[x+3][y]=p[x+3][y+4]=p[x+5][y+2]=p[x+5][y+6]=p[x+2][y+6]='+';
p[x+1][y]=p[x+2][y]=p[x+1][y+4]=p[x+2][y+4]=p[x+3][y+6]=p[x+4][y+6]='|';
p[x+4][y+1]=p[x+1][y+5]=p[x+4][y+5]='/';
p[x+1][y+1]=p[x+2][y+1]=p[x+1][y+2]=p[x+2][y+2]=p[x+1][y+3]=p[x+2][y+3]=p[x+4][y+2]=p[x+4][y+3]=p[x+4][y+4]=p[x+2][y+5]=p[x+3][y+5]=' ';
}
void print(int xx,int yy){
for(int i=xx;i>=0;i--){
for(int j=0;j<=yy;j++){
putchar(p[i][j]);
}
putchar('\n');
}
}
int main(){
while(scanf("%d%d",&n,&m)!=EOF){
int x,y;
height=0;
for(int i=0;i<n;i++){
for(int j=0;j<m;j++){
scanf("%d",&mp[i][j]);
if(mp[i][j]>height)height=mp[i][j];
}
}
memset(p,'.',sizeof(p));
prtx=prty=0;
for(int k=0;k<height;k++){
for(int i=0;i<n;i++){//draw zhengshitu
for(int j=0;j<m;j++){ if(mp[i][j]>k){
x=(n-i-1)*2+k*3;
y=j*4+(n-i-1)*2;
draw(x,y);
//printf("i=%d\t j=%d\t h=%d\t\tx:%d\ty:%d\t\n",i,j,k,x,y);
if(prtx < x+5)prtx=x+5;
if(prty < y+6)prty=y+6;
}
}
}
}
print(prtx,prty);
//output
prtx=prty=0;
memset(p,'.',sizeof(p));
//printf("%d%dn",n,m);
for(int k=0;k<height;k++){ for(int i=m-1;i>=0;i--){//draw zhengshitu
for(int j=0;j<n;j++){ if(mp[j][i]>k){
//printf("i:%dtj:%dth:%dtn",i,j,k);
x=i*2+k*3;
y=j*4+i*2;
draw(x,y);
//printf("i=%d\t j=%d\t h=%d\t\tx:%d\ty:%d\t\n",i,j,k,x,y);
if(prtx < x+5)prtx=x+5;
if(prty < y+6)prty=y+6;
}
}
}
}
print(prtx,prty);
//output
}
return 0;
}