hdu 1510

White Rectangles

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

[align=left]Problem Description[/align]
You are given a chessboard made up of N squares by N squares with equal size. Some of the squares are colored black, and the others are colored white. Please write a program to calculate the number of rectangles which are completely
made up of white squares.

 

[align=left]Input[/align]
There are multiple test cases. Each test case begins with an integer N (1 <= N <= 100), the board size. The following N lines, each with N characters, have only two valid character values:

# - (sharp) representing a black square;

. - (point) representing a white square.

Process to the end of file.

 

[align=left]Output[/align]
For each test case in the input, your program must output the number of white rectangles, as shown in the sample output.

 

[align=left]Sample Input[/align]

2
.#
..
4
..#.
##.#
.#..
.#.#

 

[align=left]Sample Output[/align]

5
15

 

解题思路：这道题我一开始的想法就是先转换成01矩阵，然后再用二维树状数组判断是否能够围成一个矩形，结果TLE了。。

TLE:

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;

char map[100][100];
int n,tree[400][400];

int lowbit(int x)
{
return x & (-x);
}

void update(int x,int y,int d)
{
for(int i = x; i <= n; i += lowbit(i))
for(int j = y; j <= n; j += lowbit(j))
tree[i][j] += d;
}

int getsum(int x,int y)
{
int sum = 0;
for(int i = x; i > 0; i -= lowbit(i))
for(int j = y; j > 0; j -= lowbit(j))
sum += tree[i][j];
return sum;
}

int main()
{
while(scanf("%d",&n)!=EOF)
{
for(int i = 1; i <= n; i++)
{
getchar();
scanf("%s",map[i]+1);
}
memset(tree,0,sizeof(tree));
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
if(map[i][j] == '.')
update(i,j,1);
int ans = 0;
for(int i = 1; i <= n; i++)
for(int j = 1; j <= n; j++)
{
for(int row = i; row >= 1; row--)
for(int col = j; col >= 1; col--)
{
int sum = getsum(i,j) - getsum(i,col-1) - getsum(row-1,j) + getsum(row-1,col-1);
int s = (i - row + 1) * (j - col + 1);
if(s == sum) ans++;
else break;
}
}
printf("%d\n",ans);
}
return 0;
}

看了别人的解题报告：(有规律可循)

.   .   .   .      1  1  1  1

 .   .   .   .      2  2  2  2

 .   .   .   .      3  3  3  3

 .   .   .   .      4  4  4  4

以这种情况为例：一共有

1*4+2*4+3*4+4*4+（1）+（1+1）+（1+1+1）+（2）+（2+2）+（2+2+2）+(3)+（3+3）+(3+3+3)+(4)+(4+4)+(4+4+4)

AC:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
const int maxn=110;
char A[maxn][maxn];
int a[maxn][maxn];
int main()
{
int T;
int sum;
while (cin>>T){
getchar();
sum=0;
memset(a,0,sizeof(a));
for(int i=0;i<T;i++){
gets(A[i]);
int l=strlen(A[i]);
for(int j=0;j<l;j++){
if(A[i][j]=='#') a[i][j]=0;
else if(!i) a[i][j]=1;
else a[i][j]=a[i-1][j]+1;
sum+=a[i][j];
if(j){
int ans=a[i][j];
for(int x=j-1;x>=0;x--){
if(!a[i][x]) break;
ans=min(ans,a[i][x]);
sum+=ans;
}
}
}
}
cout<<sum<<endl;
}
return 0;
}