题目链接：hdu 3265 Posters

题目大意：就是给定N个矩形，矩形比较特殊，均被减掉了一部分，问说图形最后的覆盖面积。

解题思路：一开始做的时候以为直接做扫描线就好了，一个做加的一个做减的，后来写完样例都跑不出来，还是对扫描线理解的不够深刻，因为扫描线没有pushdown的操作，因为它肯定对于每段区间有加有减，那么如果碰到一开始就是减的，就没法做了。
正解是将一个图形差分成至多4个小的矩形表示，然后直接扫描线。

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int maxn = 50005;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], v[maxn << 2], s[maxn << 2];

inline void pushup (int u) {
    if (v[u])
        s[u] = rc[u] - lc[u] + 1;
    else if (rc[u] == lc[u])
        s[u] = 0;
    else
        s[u] = s[lson(u)] + s[rson(u)];
}

inline void maintain (int u, int d) {
    v[u] += d;
    pushup(u);
}

void build (int u, int l, int r) {
    lc[u] = l;
    rc[u] = r;
    s[u] = v[u] = 0;

    if (l == r)
        return ;

    int mid = (l + r) / 2;
    build(lson(u), l, mid);
    build(rson(u), mid + 1, r);
    pushup(u);
}

void modify (int u, int l, int r, int d) {
    if (l <= lc[u] && rc[u] <= r) {
        maintain(u, d);
        return;
    }

    int mid = (lc[u] + rc[u]) / 2;
    if (l <= mid)
        modify(lson(u), l, r, d);
    if (r > mid)
        modify(rson(u), l, r, d);
    pushup(u);
}

struct Seg {
    int x, l, r, d;
    Seg (int x = 0, int l = 0, int r = 0, int d = 0) {
        this->x = x;
        this->l = l;
        this->r = r;
        this->d = d;
    }
};

typedef long long ll;

int N;
vector<Seg> vec;

inline bool cmp (const Seg& a, const Seg& b) {
        return a.x < b.x;
}

inline void add (int x1, int y1, int x2, int y2) {
    if (y1 == y2 || x1 == x2)
        return;
    vec.push_back(Seg(x1, y1, y2 - 1, 1));
    vec.push_back(Seg(x2, y1, y2 - 1, -1));
}

void init () {
    int x1, x2, x3, x4;
    int y1, y2, y3, y4;

    scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
    scanf("%d%d%d%d", &x3, &y3, &x4, &y4);
    add (x1, y1, x2, y3);
    add (x1, y3, x3, y4);
    add (x4, y3, x2, y4);
    add (x1, y4, x2, y2);
}

int main () {
    while (scanf("%d", &N) == 1 && N) {
        build(1, 0, 50000);
        vec.clear();
        for (int i = 0; i < N; i++)
            init();
        sort(vec.begin(), vec.end(), cmp);

        ll ans = 0;
        for (int i = 0; i < vec.size(); i++) {
            modify(1, vec[i].l, vec[i].r, vec[i].d);
            if (i != vec.size() - 1)
                ans += 1LL * s[1] * (vec[i+1].x - vec[i].x);
        }
        printf("%I64d\n", ans);
    }
    return 0;
}

hdu 3265 Posters(扫描线)