UVA 10562 Undraw the Trees（多叉树的dfs）

题意：将多叉树转化为括号表示法。

分析：gets读取，dfs就好了。注意，样例中一行的最后一个字母后是没有空格的。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long ll;
typedef unsigned long long llu;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const ll LL_INF = 0x3f3f3f3f3f3f3f3f;
const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 200 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char s[MAXN][MAXN];
int cnt;
void dfs(int x, int y){
    printf("%c(", s[x][y]);
    if(x == cnt){
        return;
    }
    if(s[x + 1][y] == '|'){//有子树
        int st = y;
        int et = y;
        while(s[x + 2][st - 1] == '-' && st - 1 >= 0) --st;//找左边界,且要保证下标不为负数
        while(s[x + 2][et] == '-') ++et;//找右边界
        for(int j = st; j < et && s[x + 3][j] != '\0'; ++j){//横线的右边界下不一定有结点，所以s[x + 3][j] != '\0'
            if(!isspace(s[x + 3][j])){
                dfs(x + 3, j);
                printf(")");
            }
        }
    }
}
int main(){
    //freopen("in.txt", "r", stdin);
    int T;
    scanf("%d", &T);
    getchar();
    while(T--){
        memset(s, 0, sizeof s);
        cnt = 0;
        while(gets(s[cnt])){//gets读入
            if(s[cnt][0] == '#') break;
            ++cnt;
        }
        printf("(");
        int len = strlen(s[0]);
        for(int j = 0; j < len; ++j){
            if(cnt == 0) break;//空树
            if(!isspace(s[0][j])){//是结点
                dfs(0, j);
                printf(")");
            }
        }
        printf(")\n");
    }
    return 0;
}