杭电1312 深搜和广搜

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概述

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

题意：

点的可以经过 #的不可以经过 @为起点

巧妙的是可以把.转换成# 当访问过之后

广搜
#include<cstdio>
#include<queue>
using namespace std;
char a[25][25];
int x,y,sum;
int dir[4][2]= {1,0,0,1,-1,0,0,-1};
struct node {
int x,y;
};
bool YES(int x0,int y0) { //xx,yy
if(x0<y&&x0>=0&&y0>=0&&y0<x)
return true;
else
return false;
}
void BFS(int x0,int y0) {
queue<node>q;
node start,endd;
start.x=x0;
start.y=y0;
q.push(start);
while(!q.empty()) {
start=q.front();
q.pop();
for(int i=0; i<4; i++) {
endd.x=start.x+dir[i][0];
endd.y=start.y+dir[i][1];
if(YES(endd.x,endd.y)&&a[endd.x][endd.y]=='.') {
a[endd.x][endd.y]='#';//将.转换成#
sum++;
q.push(endd);
}
}
}
}
int main() {
int i,j,di,dj;
while (scanf("%d%d",&x,&y)!=EOF&&x&&y) {
getchar();
for (i=0; i<y; i++) {
for (j=0; j<x; j++) {
scanf("%c",&a[i][j]);
if(a[i][j] == '@') {
di = i;
dj = j;
}
}
getchar();
}
sum=1;
BFS(di,dj);
printf("%dn",sum);
}
return 0;
}

深搜

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char map[25][25];
int vis[25][25];
int dir[4][2]={{-1,0},{0,1},{1,0},{0,-1}};
int w,h;
int dfs(int si,int sj)
{
map[si][sj]='#';
vis[si][sj]=1;
for(int i=0;i<4;i++)
{
int x=si+dir[i][0];
int y=sj+dir[i][1];
if(x<0||x>=h||y<0||y>=w||map[x][y]=='#')
continue;
dfs(x,y);
}
}
int main()
{
while(~scanf("%d%d",&w,&h)&&w&&h)
{
memset(vis,0,sizeof(vis));
for(int i=0;i<h;i++)
scanf("%s",map[i]);
bool flag=false;
for(int i=0;i<h;i++)
{
for(int j=0;j<w;j++)
{
if(map[i][j]=='@')
{
dfs(i,j);
flag=true;
break;
}
}
if(flag)
break;
}
int num=0;
for(int i=0;i<h;i++)
for(int j=0;j<w;j++)
{
if(vis[i][j])
num++;
}
printf("%dn",num);
}
return 0;
}