IOI1998 Picture 线段树

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1828

分析

扫描线法，线段树节点同时记录区间端点是否被覆盖，区间中有几条水平线，

累加答案时，注意线段会有覆盖，因此应加上变化量；

根据记录的水平线数量统计上水平线长度。

AC代码

#include 
#include 
#include 
#include 

using namespace std;

inline int read() {
    int num = 0, flag = 1;
    char c = getchar();
    while (c < '0' || c > '9') {
        if (c == '-') flag = -1;
        c = getchar();
    }
    while (c >= '0' && c <= '9')
        num = num * 10 + c - '0', c = getchar();
    return flag * num;
}

const int maxn = 5e3 + 5;

struct Line {
    int x, y1, y2, k;

    bool operator < (const Line& rhs) const {
        return x < rhs.x;
    }
} line[2 * maxn];

int n, a[2 * maxn];

struct SegmentTree {
    int l, r, cnt, len, lf, rf, num;
} t[8 * maxn];

inline void up(int p) {
    if (t[p].cnt) {
        t[p].len = a[t[p].r + 1] - a[t[p].l];
        t[p].lf = t[p].rf = 1, t[p].num = 2;
    }
    else if (t[p].l == t[p].r)
        t[p].len = t[p].lf = t[p].rf = t[p].num = 0;
    else {
        t[p].len = t[2 * p].len + t[2 * p + 1].len;
        t[p].lf = t[2 * p].lf, t[p].rf = t[2 * p + 1].rf;
        t[p].num = t[2 * p].num + t[2 * p + 1].num;
        if (t[2 * p].rf && t[2 * p + 1].lf) t[p].num -= 2;
    }
}

void build(int p, int l, int r) {
    t[p].l = l, t[p].r = r;
    t[p].cnt = t[p].len = t[p].lf = t[p].rf = t[p].num = 0;
    if (l == r) return;
    int mid = (l + r) >> 1;
    build(2 * p, l, mid);
    build(2 * p + 1, mid + 1, r);
}

void modify(int p, int l, int r, int d) {
    if (l <= t[p].l && t[p].r <= r) {
        t[p].cnt += d, up(p);
        return;
    }
    int mid = (t[p].l + t[p].r) >> 1;
    if (l <= mid) modify(2 * p, l, r, d);
    if (r > mid) modify(2 * p + 1, l, r, d);
    up(p);
}

int main() {
    while (scanf("%d", &n) == 1) {
        for (int i = 1; i <= n; ++i) {
            int x1 = read(), y1 = read(), x2 = read(), y2 = read();
            line[i] = (Line){x1, y1, y2, 1};
            line[i + n] = (Line){x2, y1, y2, -1};
            a[i] = y1, a[i + n] = y2;
        }
        sort(line + 1, line + 2 * n + 1);
        sort(a + 1, a + 2 * n + 1);
        a[0] = unique(a + 1, a + 2 * n + 1) - a - 1;
        build(1, 1, a[0] - 1);
        int ans = 0, last = 0;
        for (int i = 1; i <= 2 * n; ++i) {
            int y1 = lower_bound(a + 1, a + a[0] + 1, line[i].y1) - a;
            int y2 = lower_bound(a + 1, a + a[0] + 1, line[i].y2) - a;
            modify(1, y1, y2 - 1, line[i].k);
            ans += abs(t[1].len - last) + t[1].num * (line[i + 1].x - line[i].x);
            last = t[1].len;
        }
        printf("%d\n", ans);
    }
    return 0;
}