Satellite Photographs

Satellite Photographs

时间: 1ms        内存:128M

描述:

Farmer John purchased satellite photos of W x H pixels of his farm (1 <= W <= 80, 1 <= H <= 1000) and wishes to determine the largest 'contiguous' (connected) pasture. Pastures are contiguous when any pair of pixels in a pasture can be connected by traversing adjacent vertical or horizontal pixels that are part of the pasture. (It is easy to create pastures with very strange shapes, even circles that surround other circles.)

Each photo has been digitally enhanced to show pasture area as an asterisk ('*') and non-pasture area as a period ('.'). Here is a 10 x 5 sample satellite photo:

..*.....**
.**..*****
.*...*....
..****.***
..****.***

This photo shows three contiguous pastures of 4, 16, and 6 pixels. Help FJ find the largest contiguous pasture in each of his satellite photos.

输入:

* Line 1: Two space-separated integers: W and H
* Lines 2..H+1: Each line contains W "*" or "." characters representing one raster line of a satellite photograph.

 

输出:

* Line 1: The size of the largest contiguous field in the satellite photo.

示例输入:

10 5
..*.....**
.**..*****
.*...*....
..****.***
..****.***

示例输出:

16

提示:

参考答案(内存最优[1092]):

#include<stdio.h>

int r[] = {0, 0, 1, -1};
int c[] = {1, -1, 0, 0};

int maxi(int a, int b)
{
    if(a > b) return a;
    return b;
}

int isValid(char graph[][100],int row, int col, int maxRow, int maxCol)
{
    if(row < 0 || row >= maxRow) return 0;
    if(col < 0 || col >= maxCol) return 0;
    if(graph[row][col] != '*') return 0;
    return 1;
}

int dfs(char graph[][100], int row, int col, int maxRow, int maxCol)
{
    if(!isValid(graph, row, col, maxRow, maxCol))
        return 0;
    graph[row][col] = '#';
    int ret = 1 ,i;
    for(i=0; i<4; i++)
        ret += dfs(graph, row+r[i], col+c[i], maxRow, maxCol);
    return ret;
}

int main(void)
{
    int row, col;
    while(~scanf("%d%d",&col,&row))
    {
        int answer = 0;
        int i,j;
        char graph[1100][100];
        for( i=0; i<row; i++)
        {
            scanf("%*c");   /* 掉过换行 */
            for( j=0; j<col; j++)
                scanf("%c",&graph[i][j]);

        }
        for( i=0; i<row; i++)
            for( j=0; j<col; j++)
                if(graph[i][j] == '*')
                    answer = maxi(answer, dfs(graph, i, j, row, col));
        printf("%d\n",answer);
    }
}

参考答案(时间最优[0]):

#include<iostream>
using namespace std;

int r[] = {0, 0, 1, -1};
int c[] = {1, -1, 0, 0};

int maxi(int a, int b) {
	if(a > b) return a;
	return b;
}

int dfs(char graph[][100], int row, int col, int maxRow, int maxCol) {
	if(row < 0 || row >= maxRow) return 0;
	if(col < 0 || col >= maxCol) return 0;
	if(graph[row][col] != '*') return 0;
	
	graph[row][col] = '#';
	int ret = 1;

	for(int i=0; i<4; i++) ret += dfs(graph, row+r[i], col+c[i], maxRow, maxCol);

	return ret;
}


int main(void) {
	int row, col;
	while(cin >> col >> row) {
		int answer = 0;
		char graph[1100][100];
		for(int i=0; i<row; i++)
			for(int j=0; j<col; j++) cin >> graph[i][j];
		for(int i=0; i<row; i++) {
			for(int j=0; j<col; j++) {
				if(graph[i][j] == '*') {
					answer = maxi(answer, dfs(graph, i, j, row, col));
				}
			}
		}
		cout << answer << endl;
	}
}

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