Codeforces 1027F Session in BSU - 并查集

题目传送门

　　传送门I

　　传送门II

　　传送门III

题目大意

　　有$n​$门科目有考试，第$i​$门科目有两场考试，时间分别在$a_i, b_i\ \ (a_i < b_i)​$，要求每门科目至少参加一场考试，不能在同一个时间参加两场考试，问最后参加的考试最早的时间是什么。

　　这几天，我怎么做的都是水题Emm....

　　考虑先将$a_i, b_i$离散化。

　　对于每一门考试，在$a_i, b_i$间连一条无向边。

　　对于每个连通块，讨论：

• 如果边数大于点数，显然无解
• 如果边数等于点数，那么答案必须大于等于点权中的最大值。
• 如果边数小于点数，那么最大不可用的时候会产生若干个基环树，所以答案必须大于等于点权中的次大值。

　　并查集随便维护一下就行了。

Code

 /**
  * Codeforces
  * Problem#1027F
  * Accepted
  * Time: 873ms
  * Memory: 63000k
  */
 #include <algorithm>
 #include <iostream>
 #include <cstdlib>
 #include <cstdio>
 #include <vector>
 using namespace std;
 typedef bool boolean;

 template <typename T>
 void pfill(T* pst, const T* ped, T val) {
     for ( ; pst != ped; *(pst++) = val);
 }

 typedef class Input {
     protected:
         const static int limit = ;
         FILE* file; 

         int ss, st;
         char buf[limit];
     public:

         Input():file(NULL)    {    };
         Input(FILE* file):file(file) {    }

         void open(FILE *file) {
             this->file = file;
         }

         void open(const char* filename) {
             file = fopen(filename, "r");
         }

         char pick() {
             if (ss == st)
                 st = fread(buf, , limit, file), ss = ;//, cerr << "str: " << buf << "ed " << st << endl;
             return buf[ss++];
         }
 }Input;

 #define digit(_x) ((_x) >= '0' && (_x) <= '9')

 Input& operator >> (Input& in, unsigned& u) {
     char x;
     while (~(x = in.pick()) && !digit(x));
     for (u = x - ''; ~(x = in.pick()) && digit(x); u = u *  + x - '');
     return in;
 }

 Input& operator >> (Input& in, unsigned long long& u) {
     char x;
     while (~(x = in.pick()) && !digit(x));
     for (u = x - ''; ~(x = in.pick()) && digit(x); u = u *  + x - '');
     return in;
 }

 Input& operator >> (Input& in, int& u) {
     char x;
     while (~(x = in.pick()) && !digit(x) && x != '-');
     int aflag = ((x == '-') ? (x = in.pick(), -) : ());
     for (u = x - ''; ~(x = in.pick()) && digit(x); u = u *  + x - '');
     u *= aflag;
     return in;
 }

 Input& operator >> (Input& in, long long& u) {
     char x;
     while (~(x = in.pick()) && !digit(x) && x != '-');
     int aflag = ((x == '-') ? (x = in.pick(), -) : ());
     for (u = x - ''; ~(x = in.pick()) && digit(x); u = u *  + x - '');
     u *= aflag;
     return in;
 }

 Input in (stdin);

 #define pii pair<int, int>

 int n, m;
 int *uf;
 int *es;
 int *mx, *smx;
 pii* ps;
 vector<int> br;

 int find(int x) {
     return (uf[x] == x) ? (x) : (uf[x] = find(uf[x]));
 }

 inline void init() {
     in >> n;
     ps = new pii[(n + )];
     for (int i = ; i <= n; i++)
         in >> ps[i].first >> ps[i].second;
 }

 inline void descrete() {
     for (int i = ; i <= n; i++) {
         br.push_back(ps[i].first);
         br.push_back(ps[i].second);
     }
     sort(br.begin(), br.end());
     m = unique(br.begin(), br.end()) - br.begin();
     for (int i = ; i <= n; i++) {
         ps[i].first = lower_bound(br.begin(), br.begin() + m, ps[i].first) - br.begin();
         ps[i].second = lower_bound(br.begin(), br.begin() + m, ps[i].second) - br.begin();
     }
 }

 void addEdge(int u, int v) {
     if (find(u) != find(v)) {
         u = find(u), v = find(v);
         es[u] += es[v];
         if (mx[u] < mx[v])
             smx[u] = mx[u], mx[u] = mx[v];
         else if (smx[u] < mx[v])
             smx[u] = mx[v];
         smx[u] = ((smx[u] > smx[v]) ? (smx[u]) : (smx[v]));
         uf[v] = u;
     }
     es[find(u)]++;
 }

 int res = -;
 inline void solve() {
     uf = new int[m];
     es = new int[m];
     mx = new int[m];
     smx = new int[m];

     for (int i = ; i < m; i++)
         uf[i] = i, es[i] = -, mx[i] = i, smx[i] = -;
     for (int i = ; i <= n; i++)
         addEdge(ps[i].first, ps[i].second);
     for (int i = ; i < m; i++)
         if (find(i) == i) {
             if (es[i] > ) {
                 res = -;
                 break;
             }
             if (!es[i])
                 res = max(res, mx[i]);
             else
                 res = max(res, smx[i]);
         }

     if (res == -)
         puts("-1");
     else
         printf("%d\n", br[res]);
 }

 int main() {
     init();
     descrete();
     solve();
     return ;
 }