USACO section3.3 Home on the Range(压缩+枚举)

Home on the Range

Farmer John grazes his cows on a large, square field N (2 <= N <= 250) miles on a side (because, for some reason, his cows will only graze on precisely square land segments). Regrettably, the cows have ravaged some of the land (always in 1 mile square increments).
FJ needs to map the remaining squares (at least 2x2 on a side) on which his cows can graze (in these larger squares, no 1x1 mile segments are ravaged).

Your task is to count up all the various square grazing areas within the supplied dataset and report the number of square grazing areas (of sizes >= 2x2) remaining. Of course, grazing areas may overlap for purposes of this report.

PROGRAM NAME: range

INPUT FORMAT

Line 1:	N, the number of miles on each side of the field.
Line 2..N+1:	N characters with no spaces. 0 represents "ravaged for that block; 1 represents "ready to eat".

SAMPLE INPUT (file range.in)

6
101111
001111
111111
001111
101101
111001

OUTPUT FORMAT

Potentially several lines with the size of the square and the number of such squares that exist. Order them in ascending order from smallest to largest size.

SAMPLE OUTPUT (file range.out)

2 10
3 4
4 1

思路：这题是二维的，我的思路是吧它压成一维的，那只要判断连续的n个都满足题意，那N*N的也就满足题意。

怎么压缩呢？很简单，就是当前层是1那么就加上之前的连续1的个数，如果是0就不管它。

然后枚举矩阵大小从2到m就行了。

for(int i=m-2;i>=0;i--)

{

for(int j=0;j<m;j++)

if(map[i][j])map[i][j]+=map[i+1][j];

}

具体看代码：

/*
ID:nealgav1
LANG:C++
PROG:range
*/
#include<fstream>
#include<cstring>
using namespace std;
ifstream cin("range.in");
ofstream cout("range.out");
const int mm=255;
char s[mm][mm];
int map[mm][mm];
int num[mm];
int m;
int main()
{
while(cin>>m)
{
for(int i=0;i<m;i++)
for(int j=0;j<m;j++)
{cin>>s[i][j];map[i][j]=s[i][j]=='1'?1:0;}
for(int i=m-2;i>=0;i--)
{
for(int j=0;j<m;j++)
if(map[i][j])map[i][j]+=map[i+1][j];
}
memset(num,0,sizeof(num));
int pos;
for(int k=2;k<=m;k++)
for(int i=0;i<m;i++)
{ pos=0;
for(int j=0;j<m;j++)
{
if(map[i][j]>=k)pos++;
else pos=0;
if(pos>=k)num[k]++;
}
}
for(int i=2;i<=m;i++)
if(num[i])cout<<i<<" "<<num[i]<<"\n";
}
}

USER: Neal Gavin Gavin [nealgav1]

TASK: range

LANG: C++

Compiling...

Compile: OK

Executing...

Test 1: TEST OK [0.000 secs, 3672 KB]

Test 2: TEST OK [0.000 secs, 3672 KB]

Test 3: TEST OK [0.000 secs, 3672 KB]

Test 4: TEST OK [0.000 secs, 3672 KB]

Test 5: TEST OK [0.000 secs, 3672 KB]

Test 6: TEST OK [0.054 secs, 3672 KB]

Test 7: TEST OK [0.054 secs, 3672 KB]

All tests OK.

YOUR PROGRAM ('range') WORKED FIRST TIME! That's fantastic

-- and a rare thing. Please accept these special automated

congratulations.

Home on the Range

Russ Cox
To count the squares, we first precompute the biggest square with lower right corner at any particular location. This is done by dynamic programming: the biggest square with lower right corner at (i, j) is the minimum of three numbers:

the number of consecutive uneaten grid units to the left
the number of consecutive uneaten grid units to the right
one plus the size of the biggest square with lower right corner at (i-1, j-1)

Once we've computed this information, counting squares is simple: go to each lower right corner and increment the counters for every square size between 2 and the biggest square ending at that corner.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

#define MAXN 250

int goodsq[MAXN][MAXN];
int bigsq[MAXN][MAXN];
int tot[MAXN+1];

int
min(int a, int b)
{
return a < b ? a : b;
}

void
main(void)
{
FILE *fin, *fout;
int i, j, k, l, n, sz;

fin = fopen("range.in", "r");
fout = fopen("range.out", "w");
assert(fin != NULL && fout != NULL);

fscanf(fin, "%d\n", &n);

for(i=0; i<n; i++) {
for(j=0; j<n; j++)
goodsq[i][j] = (getc(fin) == '1');
assert(getc(fin) == '\n');
}

/* calculate size of biggest square with lower right corner (i,j) */
for(i=0; i<n; i++) {
for(j=0; j<n; j++) {
for(k=i; k>=0; k--)
if(goodsq[k][j] == 0)
break;

for(l=j; l>=0; l--)
if(goodsq[i][l] == 0)
break;

sz = min(i-k, j-l);
if(i > 0 && j > 0)
sz = min(sz, bigsq[i-1][j-1]+1);

bigsq[i][j] = sz;
}
}

/* now just count squares */
for(i=0; i<n; i++)
for(j=0; j<n; j++)
for(k=2; k<=bigsq[i][j]; k++)
tot[k]++;

for(i=2; i<=n; i++)
if(tot[i])
fprintf(fout, "%d %d\n", i, tot[i]);

exit(0);
}

Greg Price writes:

The posted solution runs in cubic time, with quadratic storage. With a little more cleverness in the dynamic programming, the task can be accomplished with only quadratic time and linear storage, and the same amount of code and coding effort. Instead of
running back along the rows and columns from each square, we use the biggest-square values immediately to the west and north, so that each non-ravaged square's biggest-square value is one more than the minimum of the values to the west, north, and northwest.
This saves time, bringing us from cubic to quadratic time.

Another improvement, which saves space and perhaps cleans up the code marginally, is to keep track of the number of squares of a given size as we go along. This obviates the need to keep a quadratic-size matrix of biggest-square values, because we only need
the most recent row for continuing the computation. As for "ravaged" values, we only use each one once, all in order; we can just read those as we need them.

#include <fstream.h>

ifstream fin("range.in");
ofstream fout("range.out");

const unsigned short maxn = 250 + 5;

unsigned short n;
char fieldpr;
unsigned short sq[maxn]; // biggest-square values
unsigned short sqpr;
unsigned short numsq[maxn]; // number of squares of each size

unsigned short
min3(unsigned short a, unsigned short b, unsigned short c)
{
if ((a <= b) && (a <= c))
return a;
else
return (b <= c) ? b : c;
}

void
main()
{
unsigned short r, c;
unsigned short i;
unsigned short tmp;

fin >> n;

for (c = 1; c <= n; c++)
sq[c] = 0;

for (i = 2; i <= n; i++)
numsq[i] = 0;

for (r = 1; r <= n; r++)
{
sqpr = 0;
sq[0] = 0;
for (c = 1; c <= n; c++)
{
fin >> fieldpr;
if (!(fieldpr - '0'))
{
sqpr = sq[c];
sq[c] = 0;
continue;
}

// Only three values needed.
tmp = 1 + min3(sq[c-1], sqpr, sq[c]);
sqpr = sq[c];
sq[c] = tmp;

// Only count maximal squares, for now.
if (sq[c] >= 2)
numsq[ sq[c] ]++;
}
}

// Count all squares, not just maximal.
for (i = n-1; i >= 2; i--)
numsq[i] += numsq[i+1];

for (i = 2; i <= n && numsq[i]; i++)
fout << i << ' ' << numsq[i] << endl;
}

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