HDU–2952 — Counting Sheep [廣搜]

 

Counting Sheep

 

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1706    Accepted Submission(s): 1104

Problem DescriptionA while ago I had trouble sleeping. I used to lie awake, staring at the ceiling, for hours and hours. Then one day my grandmother suggested I tried counting sheep after I'd gone to bed. As always when my grandmother suggests things,
I decided to try it out. The only problem was, there were no sheep around to be counted when I went to bed.

Creative as I am, that wasn't going to stop me. I sat down and wrote a computer program that made a grid of characters, where # represents a sheep, while . is grass (or whatever you like, just not sheep). To make the counting a little more interesting, I also
decided I wanted to count flocks of sheep instead of single sheep. Two sheep are in the same flock if they share a common side (up, down, right or left). Also, if sheep A is in the same flock as sheep B, and sheep B is in the same flock as sheep C, then sheeps
A and C are in the same flock.

Now, I've got a new problem. Though counting these sheep actually helps me fall asleep, I find that it is extremely boring. To solve this, I've decided I need another computer program that does the counting for me. Then I'll be able to just start both these
programs before I go to bed, and I'll sleep tight until the morning without any disturbances. I need you to write this program for me.

 InputThe first line of input contains a single number T, the number of test cases to follow.

Each test case begins with a line containing two numbers, H and W, the height and width of the sheep grid. Then follows H lines, each containing W characters (either # or .), describing that part of the grid.

 OutputFor each test case, output a line containing a single number, the amount of sheep flock son that grid according to the rules stated in the problem description.

Notes and Constraints
0 < T <= 100
0 < H,W <= 100 Sample Input

24 4#.#..#.##.##.#.#3 5###.#..#..#.###

 Sample Output

63

 

 

Code:

 

人家都用深搜, 為了練廣搜,  哼哼!

可是萬惡的Memory Limit Exceeded ! 是Memory ! MLE!  KILL IT !

找到#號的地方把這一塊的#號都標誌, 原理跟深搜一樣一樣的~

 

#include<stdio.h>#include<queue>#include<string.h>#include<iostream>using namespace std;typedef struct point{    int x,y;}point;char sign[100][100];int n,m;int dir[4][2] = {-1,0,0,1,1,0,0,-1};void bfs(int a,int b){    queue<point>q;       point p,head;    p.x = a; p.y = b;            q.push(p);          int i;    sign[a][b] = 1;       while(!q.empty()){              head = q.front();        q.pop();         for(i=0;i<4;i++)//四個方向        {            point next;            next.x = head.x + dir[i][0];            next.y = head.y + dir[i][1];            if(next.x>=0 && next.x<n && next.y>=0 && next.y<m && !sign[next.x][next.y]){                                                q.push(next);                sign[next.x][next.y] =1;//就是這裡!!!改變標誌位不能在出隊的時候,否則一判斷重複就MLE了啊,真是傳奇!            }                  }        }}int main(){    int t,i,j,count;    char str[105];    scanf("%d",&t);    while(t--)    {            char str[105];        scanf("%d%d",&n,&m);        for(i=0;i<n;i++)        {            scanf("%s",str);            for(j=0;j<m;j++)            {                if(str[j]=='#') sign[i][j] = 0;                else sign[i][j] = 1;            }            }        count = 0;        for(i=0;i<n;i++)        {            for(j=0;j<m;j++)            {                if(sign[i][j]==0)                {                    bfs(i,j);                    count ++;                }                           }        }        printf("%d\n",count);            }    return 0;}