uva 657

很简单的题，就是题意不懂……！

就是判断每个'*'区域内‘X’区域块的个数

WA了好多次，就是太差了；

1.结果排序输出

2.因为是骰子所以不再1-6范围内的数字要舍弃

3.格式要求要空一行……

4.因为碰到X就清零了，所以

XXX*X
XXX*X
.....
X***X
XX***

把X清零之后，在原来的*区域内的dfs就进行不下去了，就wa了……233333333

代码：

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <stack>
#include <cctype>
#include <string>
#include <malloc.h>
#include <queue>
#include <map>

using namespace std;
const int INF = 0xffffff;
const double Pi =  * atan();
const int Maxn =  + ;
//int dir2[8][2] = {{-1,0},{0,-1},{-1,1},{1,-1},{-1,-1},{1,0},{0,1},{1,1}};
int dr[] = {,,,-};
int dc[] = {,,-,};
int h,w;
char graph[][];
int a[];
int cnt;
int num;

void dfs2(int r,int c){
    graph[r][c] = '#';
    for(int i = ;i < ;i++){
        int xx = r + dr[i];
        int yy = c + dc[i];
        if(xx > - && yy > - && xx < h && yy < w){
            if(graph[xx][yy] == 'X'){
                dfs2(xx,yy);
            }
        }
    }
}

void dfs(int r,int c){
    if(graph[r][c] == 'X'){
        dfs2(r,c);
        num++;
    }
    graph[r][c] = ' ';
    for(int i = ;i < ;i++){
        int xx = r + dr[i];
        int yy = c + dc[i];
        if(xx > - && yy > - && xx < h && yy < w){
            if(graph[xx][yy] == '*' || graph[xx][yy] == 'X' || graph[xx][yy] == '#')
                dfs(xx,yy);
        }
    }
}

int cmp(const void * a,const void * b){
    return *((int *)a) - *((int *)b);
}

int main()
{
#ifndef ONLINE_JUDGE
    freopen("inpt.txt","r",stdin);
#endif
    int cas = ;
    while(cin >> w >> h){
        if(!w && !h)
            break;
        memset(a,,sizeof(a));
        cnt = ;
        for(int i = ;i < h;i++)
            cin >> graph[i];
        for(int i = ;i < h;i++){
            for(int j = ;j < w;j++){
                if(graph[i][j] == '*' || graph[i][j] == 'X'){
                    num = ;
                    dfs(i,j);
                    if(num >  && num < )
                        a[cnt++] = num;
                }
            }
        }
        qsort(a,cnt,sizeof(int),cmp);
        cout << "Throw " << ++cas << endl;
        for(int i = ;i < cnt-;i++)
            cout << a[i] << " ";
        cout << a[cnt-] << endl << endl;
    }
    return ;
}

测试用例：

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

XXX*X
XXX*X
.....
X***X
XX***

..........
..X**.*X..
..........
...*X*X...
..........

..........
..X....X..
..........
...*X*X...
..........

..........
..X....X..
..X....X..
..XXXXXX..
..........

..........
..X*X.....
..*X*.....
..X*X.....
..........

..........
..X*X.....
..*X**....
..X*X*....
..........

XXXXX
XXXXX
XXXXX
XXXXX
XXXXX

..............................
..............................
...............*..............
...*****......****............
...*X***.....**X***...........
...*****....***X**............
...***X*.....****.............
...*****.......*..............
..............................
........***........******.....
.......**X****.....*X**X*.....
......*******......******.....
.....****X**.......*X**X*.....
........***........******.....
..............................

..........
.XXX......
....XXX...
.......XXX
....XXX...
*X*X......

XXXXX*
.....X
.....X
.....X
.....X
.....X

XXXXX.
.....X
.....X
.....X
.....X
.....X

XXXXX.
....*X
.....X
.....X
.....X
.....X

.....X*X*X*X*X*X***...........
.X......................X.....
...............*.........X....
...X****......****........X...
...*X*.*.....**X***X..........
...*.X......***X**.....XXX....
...*.*X*.....****........X....
...***.X.......*.........X....
..............................
.......X***.............*..***
......******X****.....*X**X*..
..***********......**.*.*.....
.....****X**.......*X**X*.....
........***........*....*.....
........***.********..........
 

ans：

Throw 

Throw 

Throw 

Throw 

Throw 

Throw 

Throw 

Throw 

Throw 

Throw 

Throw 

Throw 

Throw 

Throw