Personal programming language Gym

题目链接：https://vjudge.net/contest/181019#problem/B

题意：定义函数，若 def a = value则a就是value；若def a = valuea  def b with a = valueb则b为valueb valuea；若def a = valuea   def b with a = valueb  def c with b with a = valuec 则因为b中包含a，则在这个串总去掉前面出现的a，只保留最后一个出现的a，即去掉b中的a，保留a中的a，故c为valuec valueb valuea。输入一个串s，求s代表的值，输入保证s一定是def过的。

思路：将每个串与其with的串看做两个点连接起来，查询串s时，已串s为起点，在与其相连的其他串（与其相连的点）中dfs按要求输出即可，vis数组标记一个串是否输出过了。

代码如下:

 

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<cstdlib>#include<sstream>#include<deque>#include<stack>#include<set>#include<map>using namespace std;typedef long long ll;typedef unsigned long long ull;const double eps = 1e-6;const int  maxn = 1e5 + 20;const int  maxt = 300 + 10;const int mod = 10;const int dx[] = {1, -1, 0, 0};const int dy[] = {0, 0, -1, 1};const int Dis[] = {-1, 1, -5, 5};const int inf = 0x3f3f3f3f;const int MOD = 1000;int n, m, k;map<string, int> id;map<int, string> value;bool vis[maxn];vector<int> g[maxn];vector<string> ans;void dfs(int u){    int len = g[u].size();    int v;    for(int i = len - 1; i >= 0; --i){//注意一定要倒着遍历，不然def a = 1 def b = 2 def c with b with a = 3的结果会变成3 1 2；应该是3 2 1        v = g[u][i];        if(!vis[v]){            dfs(v);            vis[v] = true;        }    }    if(!vis[u]){        ans.push_back(value[u]);        vis[u] = true;    }}int main(){    scanf("%d", &n);    getchar();    int cnt = 0;    bool ok;    string str, s;    for(int i = 1; i <= n; ++i){        getline(cin, str);//一行一行的读        stringstream ss(str);        int u;        while(ss >> s){            if(s == "def"){                ss >> s;                if(id.find(s) == id.end()){                    id[s] = cnt++;                }                u = id[s];            }            else if(s == "with") continue;            else if(s == "="){                ss >> s;                value[u] = s;            }            else{                if(id.find(s) == id.end()) id[s] = cnt++;                g[u].push_back(id[s]);//连边            }        }    }    cin >> s;    memset(vis, false, sizeof vis);    ans.clear();    dfs(id[s]);//以要查询的串对应的点为起点进行dfs，遍历结果存在ans中    int len = ans.size();    for(int i = len - 1; i >= 0; --i){//倒叙输出        cout << ans[i];        if(i) printf(" ");    }    cout << endl;    return 0;}/*6def f1 = worlddef f2 = hellodef f3 with f2 with f1 = saydef f4 = pleasedef f5 with f3 with f4 = dontdef f6 with f3 with f2 = rather*/