UVA - 10562 Undraw the Trees

来源：互联网发布：js页面刷新计数器编辑：程序博客网时间：2024/06/05 00:27

题目链接：http://vjudge.net/problem/UVA-10562

题意：将一棵多叉树转化成题目所要求的括号表示法。

分析：不用建树，将整棵树放在二维数组里一点点读取节点即可。题目不难，就是不太好解释。

#include<cstdio>#include<cstdlib>#include<string>#include<cstring>#include<iostream>#include<algorithm>#include<set>#include<map>#include<vector>#include<list>#include<deque>#include<queue>#include<cmath>using namespace std;const int maxn = 210;char s[maxn][maxn];char ans[10000];int cur,d;void dfs(int deep, int left, int right){    if (deep > d) return;    for(int i = left; i <= right; i++)    {        if((i<strlen(s[deep])) && s[deep][i] != '|' && s[deep][i] != '-' && s[deep][i] != ' ' && s[deep][i] != '#')        {            ans[cur++] = s[deep][i];            ans[cur++] = '(';            if(s[deep + 1][i] == '|')            {                int l = i, r = i;                while(s[deep + 2][l] == '-') l--;                while(s[deep + 2][r] == '-') r++;                dfs(deep + 3, l + 1, r - 1);                ans[cur++] = ')';            }            else ans[cur++] = ')';        //cout << "**" << s[deep][i] << " ** " << i << endl;        }    }    return;}int main(){    int t;    scanf("%d",&t);    getchar();    while(t--)    {        d = 0;        while(gets(s[d]))//读取树存放在二维数组里        {            if(s[d][0] == '#') break;            d++;        }        cur = 0;        ans[cur++] = '(';        dfs(0, 0, strlen(s[0]) - 1);        ans[cur++] = ')';        ans[cur] = 0;        printf("%s\n",ans);    }    return 0;}/*2    A    |--------B  C   D   |   | ----- - E   F G#e|----f g#*/

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