2021ICPC台北F What a Colorful Wall (扫描线，并查集)

https://codeforces.com/gym/103443/problem/F


扫描线核心思想就是从下到上暴力跑一维，把另一维的线段化为两个点处理，若处理两个点的线段的时间复杂度是log，这样平面上n个矩形，值域为X，Y的问题，能够在nXlogn(Y)的复杂度内处理

• 离散化
• 暴力一个一个点的涂色是n^3
• 考虑用扫描线从下往上扫，那么问题就是怎么处理平行x的线段。
• 因为是颜色覆盖问题，我们倒着处理，那么一段线段[l, r]涂过色后，这段颜色不能再边，扫到这一段时候就要从l直接跳到r。可以想到并查集来实现。
• 时间复杂度约为 nXlog(Y), (并查集比logn还快一些)

#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(false) ,cin.tie(0), cout.tie(0);
//#pragma GCC optimize(3,"Ofast","inline")
#define ll long long
//#define int long long
const int N = 1e4 + 5;
const int M = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double PI = acos(-1.0);
struct node{
 int x1, y1, x2, y2, c;
} rta[N];
int lisan( vector<int> v, int x ) {
 return lower_bound(v.begin(), v.end(), x) - v.begin();
}
int p[N];
int find ( int x ) { return x == p[x] ? x : p[x] = find(p[x]); }
void merge( int a, int b ) { a = find(a), b = find(b), p[a] = b; }
int main () {
 IOS
 int n; cin >> n;
 vector<int> X, Y;
 //离散化
 for ( int i = 1; i <= n; ++ i ) {
   int x1, y1, x2, y2, c; cin >> x1 >> y1 >> x2 >> y2 >> c;
   rta[i] = { x1, y2, x2, y1, c };
   X.push_back(x1), X.push_back(x2), Y.push_back(y1), Y.push_back(y2);
 }
 sort(X.begin(), X.end()); X.erase( unique(X.begin(), X.end()), X.end());
 sort(Y.begin(), Y.end()); Y.erase ( unique ( Y.begin(), Y.end()), Y.end());
 for ( int i = 1; i <= n; ++ i ) {
   int x1 = lisan(X, rta[i].x1); int x2 = lisan(X, rta[i].x2);
   int y1 = lisan(Y, rta[i].y1); int y2 = lisan(Y, rta[i].y2); int c = rta[i].c;
   rta[i] = { x1, y1, x2, y2, c};
 }
 //离散化后的值域
 int lx = X.size(), ly = Y.size();
 vector<bool> st(n + 1);

 for ( int i = 0; i < ly; ++ i ) {
   for ( int j = 0; j < lx ; ++ j) p[j] = j; // 并查集初始化要大一
   for ( int j = n; j >= 1; -- j ) {
     int x1 = rta[j].x1, x2 = rta[j].x2, y1 = rta[j].y1, y2 = rta[j].y2, c = rta[j].c;
     if( i >= y1 && i < y2) { // 注意根据题目，得出这里比较带不带等号
       for ( int k = find(x1); k < x2; k = find(k) ) {  //跳过已经涂色的
         merge( k, k + 1 ); 
         st[c] = 1;
       }
     }
   }
 }

 cout << count( st.begin(), st.end(), 1 ) << '\n';
 return 0;
}

标签：What,x1,rta,Colorful,int,x2,扫描线,y1,y2
来源： https://www.cnblogs.com/muscletear/p/16246153.html