URAL 2019 Pair: normal and paranormal (STL栈)

题意：在一个半圆内，有2*n个点，其中有大写字母和小写字母。其中你需要连接大写字母到小写字母，其中需要保证这些连接的线段之间没有相交。

如果能够实现，将大写字母对应的小写字母的序号按序输出。

析：我把它看成一个括号序列，然后用栈解决即可。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")

#include <cstdio>

#include <string>

#include <cstdlib>

#include <cmath>

#include <iostream>

#include <cstring>

#include <set>

#include <queue>

#include <algorithm>

#include <vector>

#include <map>

#include <cctype>

#include <cmath>

#include <stack>

#include <sstream>

#define debug() puts("++++");

#define gcd(a, b) __gcd(a, b)

#define lson l,m,rt<<1

#define rson m+1,r,rt<<1|1

#define freopenr freopen("in.txt", "r", stdin)

#define freopenw freopen("out.txt", "w", stdout)

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

typedef pair<int, int> P;

const int INF = 0x3f3f3f3f;

const LL LNF = 1e16;

const double inf = 0x3f3f3f3f3f3f;

const double PI = acos(-1.0);

const double eps = 1e-8;

const int maxn = 1e5 + 10;

const int mod = 1e9 + 7;

const int dr[] = {-1, 0, 1, 0};

const int dc[] = {0, 1, 0, -1};

const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};

int n, m;

const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

inline bool is_in(int r, int c){

  return r >= 0 && r < n && c >= 0 && c < m;

}

struct Node{

  char ch;

  int id;

};

map<int, int> mp;

stack<Node> st;

int main(){

  string s;

  cin >> n >> s;

  int up = 1, lo = 1;

  for(int i = 0; i < s.size(); ++i){

    if(st.empty()){

      if(islower(s[i]))  st.push((Node){s[i], lo++});

      else st.push((Node){s[i], up++});

    }

    else if(islower(s[i])){

      if(s[i] == st.top().ch + 32)  mp[st.top().id] =  lo++, st.pop();

      else  st.push((Node){s[i], lo++});

    }

    else{

      if(s[i] + 32 == st.top().ch)  mp[up++] = st.top().id, st.pop();

      else  st.push((Node){s[i], up++});

    }

  }

  if(!st.empty()){ puts("Impossible");  return 0; }

  for(map<int, int> :: iterator it = mp.begin(); it != mp.end(); ++it)

    if(it == mp.begin())  printf("%d", it->second);

    else  printf(" %d", it->second);

  printf("\n");

  return 0;

}