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  • Gym 101190H Hard Refactoring (模拟坑题)

    题意:给定 n 个区间,让你进行合并,问你最后的区间是,如果是空集,输出 false 如果区间是是 [-32768,32767] ,则是true。

    析:进行区间合并,要注意,如果是 x >= 0 && x <= 32767 那么输出是  x >= 0,在这地方,真是错死了。。。。。。后来看了数据才知道有这个,其他的就是进行区间合并,如果第 i 个和第 j 个区间合并时,假设 x >= a && x <= b  || x >= c && x <= d,那么如果 b+1 >= c && a <= c || d+1 >= a && c <= a  就进行合并,当然对于单向的就进行一个标记左端点是-32768右端点是32767,也就是最后在输出。

    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>
    #include <list>
    #include <assert.h>
    #include <bitset>
    #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 fi first
    #define se second
    #define pb push_back
    #define sqr(x) ((x)*(x))
    #define ms(a,b) memset(a, b, sizeof a)
    #define sz size()
    #define pu push_up
    #define pd push_down
    #define cl clear()
    #define all 1,n,1
    #define FOR(x,n)  for(int i = (x); i < (n); ++i)
    #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 = 1e15;
    const double inf = 1e20;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1000 + 500;
    const LL 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{
      int small, big;
      bool flag;
    };
    
    Node a[maxn];
    
    int clac(const string &s){
      int ans = 0;
      int sig = s[0] == '-' ? -1 : 1;
      ans = s[0] == '-' ? 0 : s[0] - '0';
      for(int i = 1; i < s.sz; ++i)
        ans = ans * 10 + s[i] - '0';
      return ans * sig;
    }
    
    vector<Node> ans;
    
    int main(){
      freopen("hard.in", "r", stdin);
      freopen("hard.out", "w", stdout);
      string s;
      int idx = 0;
      while(getline(cin, s)){
        a[idx].flag = 0;
        if(s[2] == '>'){
          stringstream ss(s);
          string t;
          ss >> t;  ss >> t;
          ss >> t;
          a[idx].big = clac(t);
          if((ss >> t)) {
            if(t == "||")  a[idx].small = 32767;
            else{
              ss >> t >> t;
              ss >> t;
              a[idx].small = clac(t);
            }
          }
          else  a[idx].small = 32767;
        }
        else{
          stringstream ss(s);
          string t;
          ss >> t;  ss >> t;
          ss >> t;
          a[idx].small = clac(t);
          a[idx].big = -32768;
        }
          if(a[idx].big < -32768 || a[idx].small > 32767)  a[idx].flag = 1;
        ++idx;
      }
    
      bool ok = false;
      for(int i = 0; i < idx; ++i){
        if(a[i].small >= a[i].big){
            ok = true;
        }
      }
    
      if(!ok){  puts("false");  return 0; }
    
    
      for(int k = 0; k < 10; ++k)
        for(int i = 0; i < idx; ++i){
          if(a[i].small < a[i].big)  a[i].flag = 1;
          if(a[i].flag)  continue;
          for(int j = 0; j < idx; ++j){
            if(a[j].small < a[j].big)  a[j].flag = 1;
            if(i == j)  continue;
            if(a[j].flag)  continue;
            if(a[i].big <= a[j].small+1 && a[i].small >= a[j].small || a[j].big <= a[i].small+1 && a[i].small <= a[j].small){
              a[i].small = max(a[i].small, a[j].small);
              a[i].big = min(a[i].big, a[j].big);
              a[j].flag = 1;
            }
          }
        }
    
      for(int i = 0; i < idx; ++i)
        if(!a[i].flag)  ans.pb(a[i]);
    
    
      if(ans[0].small == 32767 && ans[0].big == -32768)  puts("true");
      else {
        for(int i = 0; i < ans.sz; ++i){
          if(ans[i].big > -32768){
            printf("x >= %d", ans[i].big);
            if(ans[i].small < 32767)  printf(" && x <= %d", ans[i].small);
          }
          else printf("x <= %d", ans[i].small);
          if(i != ans.sz-1)  printf(" ||");
          puts("");
        }
      }
    
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7450039.html
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