BZOJ4195 luoguP1955 NOI2015D1T1 程序自动分析

题意：给定n个(x_i = x_j) 或 (x_i != x_j) 的条件，问是否可能成立

BZOJ链接：http://www.lydsy.com/JudgeOnline/problem.php?id=4195

luogu链接：https://www.luogu.org/problemnew/show/P1955

离散化+并查集

离线，先把(x_i = x_j)用并查集解决，再一个一个判断不等的是否矛盾

一开始用map离散化结果TLE

emm

 #include<iostream>
 #include<cstdio>
 #include<algorithm>
 #include<cmath>
 #include<cstring>
 #include<map>
 #include<vector>

 using namespace std;

 const int MAXN = ;
 int n, kx;
 int fa[MAXN * ], size[MAXN * ], s[MAXN * ];
 vector< pair<int, int> > tf, li;

 template <typename tn> void read (tn & a) {
     tn x = , f = ;
     char c = getchar();
     while (c < '' || c > ''){ if (c == '-') f = -; c = getchar(); }
     while (c >= '' && c <= ''){ x = x *  + c - ''; c = getchar(); }
     a = f ==  ? x : -x;
 }

 int find (int x) {
     if (fa[x] == x) return x;
     return fa[x] = find(fa[x]);
 } 

 void link (int x, int y) {
     x = find(x);
     y = find(y);
     if (x == y) return;
     if (size[x] > size[y]) { link(y, x); return; }
     fa[x] = y;
     size[y] += size[x];
 }

 int ge (int x) {
     int left = , right = kx + ;
     while (left < right - ) {
         int mid = (left + right) / ;
         if (s[mid] > x) right = mid; else left = mid;
     }
     if (s[left] == x) return left; else return left + ;
 }

 int main() {
     int T;
     read(T);
     while (T--) {
         read(n);
         tf.clear(); li.clear();
         for (int i = ; i <= n; ++i) {
             int x, y, k;
             read(x);
             read(y);
             read(k);
             if (k == ) tf.push_back(make_pair(x, y)); else li.push_back((make_pair(x, y)));
             s[i + i - ] = x;
             s[i + i] = y;
         }
         sort(s + , s + n + n);
         int i = ;
         kx = ;
         while (i <= n * ) {
             int j = i;
             while (j +  <= n *  && s[j + ] == s[i]) ++j;
             s[kx] = s[i];
             i = j + ;
             ++kx;
         }
         --kx;
         for (int i = ; i <= n + n; ++i) fa[i] = i, size[i] = ;
         for (int i = ; i < li.size(); ++i) {
             int x, y;
             x = li[i].first;
             y = li[i].second;
             x = ge(x);
             y = ge(y);
             link(x, y);
         }
         bool f = ;
         for (int i = ; i < tf.size(); ++i) {
             int x, y;
             x = tf[i].first;
             y = tf[i].second;
             x = ge(x);
             y = ge(y);
             x = find(x);
             y = find(y);
             if (x == y) { f = ; break; }
         }
         if (f) cout << "YES\n"; else cout << "NO\n";
     }
     return ;
 }