HDU 5820 Lights(主席树)

题意:

$给定N\times N的网格图,N= 5\times 10^5,选中其中K\le 5\times 10^5个交叉点$
$现判断对于任意2个交叉点之间,是否至少存在一条路径,使得这个路径的每个转弯都是交叉点$

分析:


$官方题解说的劲啊,实现一下就好了$
$对于每个点(x, y)找它上面第一个点,下面第一个,以及左边第一个$
$没有就给到边界上,求这个2个矩形的内部点数就好$

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//
// Created by TaoSama on 2016-08-10
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 5e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
const int C = 5e4;
int n;
vector<int> ps[C + 5];
int root[N];
struct PersistentSegTree {
static const int M = N * 20;
int sz;
struct Node {
int ls, rs, sum;
} tree[M];
void init() {
sz = 0;
memset(&tree[0], 0, sizeof tree[0]);
}
int newNode(int rt) {
tree[++sz] = tree[rt];
return sz;
}
void update(int o, int v, int l, int r, int& rt) {
rt = newNode(rt);
tree[rt].sum += v;
if(l == r) return;
int m = l + r >> 1;
if(o <= m) update(o, v, l, m, tree[rt].ls);
else update(o, v, m + 1, r, tree[rt].rs);
}
int query(int L, int R, int l, int r, int rt) {
if(L <= l && r <= R) return tree[rt].sum;
int m = l + r >> 1, ret = 0;
if(L <= m) ret += query(L, R, l, m, tree[rt].ls);
if(R > m) ret += query(L, R, m + 1, r, tree[rt].rs);
return ret;
}
} T;
int main() {
#ifdef LOCAL
// freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
freopen("C:\\Users\\TaoSama\\Desktop\\1012.in", "r", stdin);
freopen("C:\\Users\\TaoSama\\Desktop\\out.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(0);
while(scanf("%d", &n) == 1 && n) {
vector<int> xs;
for(int i = 1; i <= C; ++i) ps[i].clear();
for(int i = 1; i <= n; ++i) {
int x, y; scanf("%d%d", &x, &y);
ps[x].push_back(y);
}
for(int i = 1; i <= C; ++i) {
auto& v = ps[i];
if(!v.size()) continue;
sort(v.begin(), v.end());
v.resize(unique(v.begin(), v.end()) - v.begin());
}
T.init();
bool ok = true;
vector<int> lastX(C + 1, 0);
for(int x = 1; x <= C && ok; ++x) {
root[x] = root[x - 1];
auto& v = ps[x];
for(int i = 0; i < v.size(); ++i) {
int y = v[i];
int preX = lastX[y];
int preY = i - 1 >= 0 ? v[i - 1] : 0;
int nxtY = i + 1 < v.size() ? v[i + 1] : C + 1;
int num1 = T.query(preY + 1, nxtY - 1, 1, C, root[preX]);
int num2 = T.query(preY + 1, nxtY - 1, 1, C, root[x - 1]);
if(num1 != num2) {ok = false; break;}
lastX[y] = x;
T.update(y, 1, 1, C, root[x]);
}
}
puts(ok ? "YES" : "NO");
}
return 0;
}

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