DFS入门题

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#output #integer #spring #input #c #each

1378: Best Grass

Time Limit:1000MS Memory Limit:65536K
Total Submit:47 Accepted:25

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Description

Bessie is planning her day of munching tender spring grass and is gazing

out upon the pasture which Farmer John has so lovingly partitioned into a

grid with R (1 <= R <= 100) rows and C (1 <= C <= 100) columns. She wishes

to count the number of grass clumps in the pasture.



Each grass clump is shown on a map as either a single '#' symbol or perhaps

two '#' symbols side-by-side (but not on a diagonal). Given a map of the

pasture, tell Bessie how many grass clumps there are.



By way of example, consider this pasture map where R=5 and C=6:



.#....

..#...

..#..#

...##.

.#....



This pasture has a total of 5 clumps: one on the first row, one that spans

the second and third row in column 2, one by itself on the third row, one

that spans columns 4 and 5 in row 4, and one more in row 5.

Input

* Line 1: Two space-separated integers: R and C



* Lines 2..R+1: Line i+1 describes row i of the field with C

        characters, each of which is a '#' or a '.'

Output

* Line 1: A single integer that is the number of grass clumps Bessie

        can munch

Sample Input

5 6

.#....

..#...

..#..#

...##.

.#....

Sample Output

Source

OPEN08

没啥多说的，就是典型的DFS

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

char map[105][105];

int j[2][4] = {0, 0, -1, 1, 1, -1, 0, 0};

int dfs(int x, int y)
{
if(map[x][y] == '#')
{
map[x][y] = '.';
for(int i = 0; i < 4; i++)
{
int xi = x + j[0][i], yj = y + j[1][i];
if(map[xi][yj] != 0)
dfs(xi, yj);
}
}
return 0;
}

int main()
{
int r, c;
memset(map, 0, sizeof(map));
scanf("%d %d", &r, &c);
int count = 0;
for(int i = 0; i < r; i++)
scanf("%s", map[i]);
for(int i = 0; i < r; i++)
for(int j = 0; j < c; j++)
if(map[i][j] == '#')
{
dfs(i, j);
count++;
}
printf("%d/n", count);
return 0;
}