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;
}

  

Gym 101190H Hard Refactoring (模拟坑题)的更多相关文章

  1. HDOJ 2317. Nasty Hacks 模拟水题

    Nasty Hacks Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Tota ...

  2. 【BZOJ-1952】城市规划 [坑题] 仙人掌DP + 最大点权独立集(改)

    1952: [Sdoi2010]城市规划 Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 73  Solved: 23[Submit][Status][ ...

  3. hdu 5455 Fang Fang 坑题

    Fang Fang Time Limit: 1 Sec Memory Limit: 256 MB 题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=5455 ...

  4. String类面试坑题

    1.面试坑题F:\SHJT\JavaWorkspace\JavaSE\workspace\day13ezra\src\cn\itcast\sh\classcode\BTStringLastIndexO ...

  5. POJ 2014:Flow Layout 模拟水题

    Flow Layout Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 3091   Accepted: 2148 Descr ...

  6. Codeforces Gym 100650C The Game of Efil 模拟+阅读题

    原题链接:http://codeforces.com/gym/100650/attachments/download/3269/20052006-acmicpc-east-central-north- ...

  7. Gym 100851E Easy Problemset (模拟题)

    Problem E. Easy ProblemsetInput file: easy.in Output file: easy.outPerhaps one of the hardest problems ...

  8. Codeforces 1082B Vova and Trophies 模拟,水题,坑 B

    Codeforces 1082B Vova and Trophies https://vjudge.net/problem/CodeForces-1082B 题目: Vova has won nn t ...

  9. hdu 5083 有坑+字符串模拟水题

    http://acm.hdu.edu.cn/showproblem.php?pid=5083 机器码和操作互相转化 注意SET还要判断末5位不为0输出Error #pragma comment(lin ...

随机推荐

  1. 第十章 Secret & Configmap (中)

    10.3 在Pod中使用Secret 10.3.1 Volume方式 apiVersion: v1 kind: Pod metaata: name: mypod spec: containers: - ...

  2. small_vector

    folly/small_vector.h folly::small_vector<T,Int=1,...> is a sequence container that implements ...

  3. 无法添加数据连接。Could not load file or assembly 'Microsoft.SqlServer.Management.Sdk.Sfc, Version=11.0.0.0

    无法添加数据连接.Could not load file or assembly 'Microsoft.SqlServer.Management.Sdk.Sfc, Version=11.0.0.0 V ...

  4. django -- 内置分页

    urls.py from django.conf.urls import url from conn_oracle import views urlpatterns = [ url(r'^page/' ...

  5. Bootstrap 与 Jquery validate 结合使用——简单实现

    首先必须引入的JS和CSS <script type="text/javascript" src="${ctx}/static/js/jquery-1.9.1.mi ...

  6. Spring cloud Eureka高可用 - Windows 7 hosts文件立即生效

    hosts 文件所在位置 c:/windows/system32/drivers/etc/hosts 左下角 搜索框 搜索 cmd 弹出命令框 输入 ipconfig /displaydns 显示所有 ...

  7. 使用copy函数输出容器中的内容

    container<type> c; 输出语句为:copy(c.begin(), c.end(), ostream_iterator<type>(cout, " &q ...

  8. 【转】SVN 与 GIT 详细对比

    git和svn的详细对比   近期就[版本管理工具是否进行切换SVN->Git]的问题进行了讨论,于是对svn和Git进行了相关研究,进而梳理出Git的特点(优.缺点),最后将Git与SVN进行 ...

  9. jQuery插件–jqueryflexselect下拉框自动提示

    原理:用户在文本框中输入一个字符(或输入字符的首字母),然后利用ajax,从后台服务器中进行检索,组装后并返回到页面,页面通过javascript进行解析,在一个层里面显示出来. 类似的效果如下: & ...

  10. 关于setTimeout()你所不知道的地方

    前言:看了这篇文章,1.注意setTimeout引用的是全部变量还是局部变量了,当直接调用外部函数方法时,实际上函数内部的变量已经变成全 局.2.提醒我防止出错的,用匿名函数不容易出错.3.setTi ...