UVA 657 The die is cast
2016-08-11 13:50
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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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