题目链接：hdu 1828 Picture

题目大意：N个矩形，求矩形周长的并。

解题思路：利用到线段数区间合并，记录有多少个连续块，还用到区间修改，每次对于一条边，除了要计算竖直方向，还要计算水平方向，而水平方向是修改后的增减量。

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

using namespace std;

const int maxn = 20005;

vector<int> pos;

#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)|1)
int lc[maxn << 2], rc[maxn << 2], set[maxn << 2];
int L[maxn << 2], R[maxn << 2], S[maxn << 2], len[maxn << 2];

inline int length(int u) {
    return rc[u] - lc[u] + 1;
}

inline void pushup (int u) {
    if (set[u]) {
        L[u] = R[u] = length(u);
        len[u] = pos[rc[u] + 1] - pos[lc[u]];
        S[u] = 1;
    } else if (lc[u] == rc[u]) {
        len[u] = 0;
        L[u] = R[u] = S[u] = 0;
    } else {
        S[u] = S[lson(u)] + S[rson(u)] + (R[lson(u)] && L[rson(u)] ? -1 : 0);
        L[u] = L[lson(u)] + (L[lson(u)] == length(lson(u)) ? L[rson(u)] : 0);
        R[u] = R[rson(u)] + (R[rson(u)] == length(rson(u)) ? R[lson(u)] : 0);
        len[u] = len[lson(u)] + len[rson(u)];
    }
}

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

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

    if (l == r) {
        maintain(u, 0);
        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 v) {
    if (l <= lc[u] && rc[u] <= r) {
        maintain(u, v);
        return;
    }

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

typedef long long ll;

struct point {
    int x1, y1;
    int x2, y2;
}p[maxn];

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

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

int N;
vector<Seg> vec;

inline int find (int x) {
    return lower_bound(pos.begin(), pos.end(), x) - pos.begin();
}

int main () {
    while (scanf("%d", &N) == 1) {
        vec.clear();
        pos.clear();

        for (int i = 0; i < N; i++) {
            scanf("%d%d%d%d", &p[i].x1, &p[i].y1, &p[i].x2, &p[i].y2);
            pos.push_back(p[i].y1);
            pos.push_back(p[i].y2);
        }
        sort(pos.begin(), pos.end());
        build(1, 0, pos.size());

        for (int i = 0; i < N; i++) {
            int l = find(p[i].y1), r = find(p[i].y2) - 1;
            vec.push_back(Seg(p[i].x1, l, r, 1));
            vec.push_back(Seg(p[i].x2, l, r, -1));
        }
        sort(vec.begin(), vec.end(), cmp);

        ll ans = 0;
        for (int i = 0; i < vec.size(); i++) {

            int tmp = len[1];
            modify(1, vec[i].l, vec[i].r, vec[i].v);
            ans += abs(tmp - len[1]);

            if (i != vec.size() - 1)
                ans += (2LL * S[1] * (vec[i+1].x - vec[i].x));
        }
        printf("%lld\n", ans);
    }
    return 0;
}

hdu 1828 Picture(线段树)