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UVA 657 The die is cast

2016-08-11 13:50 302 查看

UVA 657 The die is cast

题目大意:给出字符串 *代表骰子区域,X代表点数,若几个X相邻代表是同一个点数,求出每个骰子共多少点数,并从小到大输出

解题思路:用俩次dfs搜索,第一次搜索*的区域,然后搜索X

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
char ch[100][100];
int flag[100][100];
int sum[100];
int x;
int w, h;
int cmd(const void *a, const void *b) {
return *(int*)a - *(int*)b;
}
void dfs2(int a, int b) {
ch[a][b] = '*';
if(a - 1 >= 0 && ch[a-1][b] == 'X')
dfs2(a-1, b);
if(a + 1 < h && ch[a+1][b] == 'X')
dfs2(a+1, b);
if(b - 1 >= 0 && ch[a][b-1] == 'X')
dfs2(a, b-1);
if(b + 1 < w && ch[a][b+1] == 'X')
dfs2(a, b+1);
}

void dfs1(int a, int b) {
flag[a][b] = 1;
if(ch[a][b] == 'X') {
dfs2(a, b);
sum[x]++;
}
if(a - 1 >= 0 && ((ch[a-1][b] == '*' && flag[a-1][b] == 0)|| ch[a-1][b] == 'X'))
dfs1(a-1, b);
if(a + 1 < h && ((ch[a+1][b] == '*' && flag[a+1][b] == 0)|| ch[a+1][b] == 'X'))
dfs1(a+1, b);
if(b - 1 >= 0 && ((ch[a][b-1] == '*' && flag[a][b-1] == 0)|| ch[a][b-1] == 'X'))
dfs1(a, b-1);
if(b + 1 < w && ((ch[a][b+1] == '*' && flag[a][b+1] == 0)|| ch[a][b+1] == 'X'))
dfs1(a, b+1);
}

int main() {
int n = 0;
while(cin >> w >> h && w + h != 0) {
getchar();
n++;
x = 0;
memset(flag, 0, sizeof(flag));
for(int i = 0; i < h; i++)
gets(ch[i]);
for(int i = 0; i < h; i++)
for(int j = 0; j < w; j++) {
if((ch[i][j] == '*' || ch[i][j] == 'X') && flag[i][j] == 0) {
sum[x] = 0;
dfs1(i, j);
x++;
}
}
qsort(sum, x, sizeof(int), cmd);
printf("Throw %d\n", n);
for(int i = 0; i < x; i++) {
printf("%d", sum[i]);
if(i != x - 1)
printf(" ");
}
printf("\n\n");

}

}
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