bzoj 1818 [CQOI 2010] 内部白点

题目传送门

　　快速的列车

　　慢速的列车

题目大意

　　一个无限大的方格图内有$n$个黑点。问有多少个位置上下左右至少有一个黑点或本来是黑点。

　　扫描线是显然的。

　　考虑一下横着的线段，取它两个端点，横坐标小的地方放一个+1，大的地方放一个-1事件。

　　然后扫描，扫到的横着的线段更新，竖着的线段用树状数组求答案。

　　然后考虑这一列上原来存在的黑点有没有被统计，如果没有就加上。

Code

/**
 * bzoj
 * Problem#1818
 * Accepted
 * Time: 1824ms
 * Memory: 9776k
 */
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <vector>
#ifndef WIN32
#define Auto "%lld"
#else
#define Auto "%I64d"
#endif
using namespace std;
#define smax(a, b) (a = max(a, b))
#define smin(a, b) (a = min(a, b))
typedef bool boolean;
#define ll long long

const int N = 1e5 + 5;

typedef class IndexedTree {
    public:
        int s;
        int* ar;
        
        IndexedTree() {    }
        IndexedTree(int s):s(s) {
            ar = new int[(s + 1)];
            memset(ar, 0, sizeof(int) * (s + 1));
        }

        void add(int idx, int val) {
            for ( ; idx <= s; idx += (idx & (-idx)))
                ar[idx] += val;
        }

        int query(int idx) {
            int rt = 0;
            for ( ; idx; idx -= (idx & (-idx)))
                rt += ar[idx];
            return rt;
        }
}IndexedTree;

typedef class Event {
    public:
        int x, y, val;

        Event(int x = 0, int y = 0, int val = 0):x(x), y(y), val(val) {    }

        boolean operator < (Event b) const {
            return x < b.x;
        }
}Event;

int n;
int xs[N], ys[N], bxs[N], bys[N];
int ls[N], rs[N], us[N], ds[N];
ll res = 0;
int tp = 0;
Event es[N << 1];
vector<int> vs[N];
IndexedTree it;

inline void init() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
        scanf("%d%d", xs + i, ys + i);
}

inline void descrete(int* ar, int* br, int n) {
    memcpy(br, ar, sizeof(int) * (n + 1));
    sort(br + 1, br + n + 1);
    for (int i = 1; i <= n; i++)
        ar[i] = lower_bound(br + 1, br + n + 1, ar[i]) - br;
}

inline void solve() {
    descrete(xs, bxs, n);
    descrete(ys, bys, n);
    
    for (int i = 1; i <= n; i++)
        ls[i] = ds[i] = N;
    for (int i = 1; i <= n; i++)
        us[i] = rs[i] = 0;

    for (int i = 1; i <= n; i++) {
        smin(ls[ys[i]], xs[i]);
        smax(rs[ys[i]], xs[i]);
        smin(ds[xs[i]], ys[i]);
        smax(us[xs[i]], ys[i]);
    }
    
    for (int i = 1; i <= n; i++)
        vs[xs[i]].push_back(ys[i]);

    for (int i = 1; i <= n; i++)
        if (ls[i] < rs[i] - 1) {
            es[++tp] = Event(ls[i], i, 1);
            es[++tp] = Event(rs[i], i, -1);
        }
    sort(es + 1, es + tp + 1);

    int pe = 1;
    it = IndexedTree(n);
    bxs[0] = -1e9 - 5;
    for (int i = 1; i <= n; i++) {
        if (bxs[i] == bxs[i - 1])    continue;
        while (pe <= tp && es[pe].x == i)
            it.add(es[pe].y, es[pe].val), pe++;
        if (ds[i] <= us[i])
            res += it.query(us[i]) - it.query(ds[i] - 1);//, cerr << bxs[i] << endl;
        for (int j = 0; j < (signed) vs[i].size(); j++)
            if (it.query(vs[i][j]) - it.query(vs[i][j] - 1) == 0)
                res++;
//        cerr << bxs[i] << " " << res << endl;
    }
    printf(Auto"
", res);
}

int main() {
    init();
    solve();
    return 0;
}