Red and Black
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 28026 Accepted Submission(s): 16916
Problem Description
There is a rectangular room, covered with square tiles. Each tile is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent tiles. But he can’t move on red tiles, he can move only on black tiles.
Write a program to count the number of black tiles which he can reach by repeating the moves described above.
Input
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.
There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.
‘.’ - a black tile
‘#’ - a red tile
‘@’ - a man on a black tile(appears exactly once in a data set)
Output
For each data set, your program should output a line which contains the number of tiles he can reach from the initial tile (including itself).
Sample Input
6 9
…#.
…#
…
…
…
…
…
#@…#
.#…#.
11 9
.#…
.#.#######.
.#.#…#.
.#.#.###.#.
.#.#…@#.#.
.#.#####.#.
.#…#.
.#########.
…
11 6
…#…#…#…
…#…#…#…
…#…#…###
…#…#…#@.
…#…#…#…
…#…#…#…
7 7
…#.#…
…#.#…
###.###
…@…
###.###
…#.#…
…#.#…
0 0
Sample Output
45
59
6
13
BFS版本：
用bfs的写法没什么特殊的，标记一下已走过的路，套用模板即可。
ac代码附上：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <ctype.h>
#include <queue>
#include <cmath>
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;
typedef long long int ll;
char mp[25][25]; // first time: int (stupid)
int sgn[25][25];
int mv[4][2] = {{-1,0},{0,1},{1,0},{0,-1}}; // up right down left
int w,h;
int si,sj;
int sum;
struct node {
int i,j;
};
bool check(int i,int j) {
if(i < 1 || i > h || j < 1 || j > w) {return false;}
if(mp[i][j] == '#') {return false;}
if(sgn[i][j] == 1) {return false;}
return true;
}
void bfs() {
memset(sgn,0,sizeof(sgn));
node cur,next;
cur.i = si;
cur.j = sj;
sgn[si][sj] = 1;
queue<node>q;
q.push(cur);
while(!q.empty()) {
cur = q.front();
q.pop();
for(int k = 0; k < 4; k++) {
next.i = cur.i + mv[k][0];
next.j = cur.j + mv[k][1];
if(check(next.i,next.j)) {
sum++;
sgn[next.i][next.j] = 1;
q.push(next);
}
}
}
}
int main() {
while(cin >> w >> h) {
if(w == 0 && h == 0) {break;}
sum = 1;
for(int i = 1; i <= h; i++) {
for(int j = 1; j <= w; j++) {
cin >> mp[i][j];
if(mp[i][j] == '@') {
si = i;
sj = j;
}
}
}
bfs();
cout << sum << endl;
}
return 0;
}
DFS版本：
记住DFS解决的是：现在要做什么
附上ac代码：
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
#include <ctype.h>
#include <queue>
#include <cmath>
#define INF 0x3f3f3f3f
#define mod 1000000007
using namespace std;
typedef long long int ll;
char mp[25][25];
int sgn[25][25];
int w,h;
int si,sj;
int sum;
void dfs(int i,int j) {
if(i < 1 || i > h || j < 1 || j > w || sgn[i][j] == 1 || mp[i][j] == '#') {return;}
sgn[i][j] = 1;
sum++;
dfs(i-1,j); // up
dfs(i,j+1); // right
dfs(i+1,j); // down
dfs(i,j-1); // left
return;
}
int main() {
while(cin >> w >> h) {
if(w == 0 && h == 0) {break;}
for(int i = 1; i <= h; i++) {
for(int j = 1; j <= w; j++) {
cin >> mp[i][j];
if(mp[i][j] == '@') {
si = i;
sj = j;
}
}
}
memset(sgn,0,sizeof(sgn));
sum = 0;
dfs(si,sj);
cout << sum << endl;
}
return 0;
}