poj 1177 Picture

http://poj.org/problem?id=1177

　　经典线段树，求矩形并的周长。把坐标排个序，然后插入线段树，求其变化的和就可以了。轻松1y~

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

using namespace std;

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define root -10001, 10001, 1

const int maxn = 40005;
struct segTree{
    int sum[maxn << 2], low[maxn << 2], high[maxn << 2], late[maxn << 2];

    void up(int rt) {
        int ls = rt << 1, rs = rt << 1 | 1;

        sum[rt] = sum[ls] + sum[rs];
        low[rt] = min(low[ls], low[rs]);
        high[rt] = max(high[ls], high[rs]);
    }
    void down(int rt) {
        if (late[rt]) {
            int ls = rt << 1, rs = rt << 1 | 1;

            late[ls] += late[rt];
            late[rs] += late[rt];
            low[ls] += late[rt];
            low[rs] += late[rt];
            high[ls] += late[rt];
            high[rs] += late[rt];
            late[rt] = 0;
        }
    }
    void build(int l, int r, int rt) {
        sum[rt] = low[rt] = high[rt] = late[rt] = 0;
        if (l == r) {
            return ;
        }
        int m = (l + r) >> 1;
        build(lson);
        build(rson);
    }
    void fix(int l, int r, int rt) {
        if (l == r || !high[rt] || low[rt]) {
            if (low[rt]) {
                sum[rt] = r - l + 1;
            } else if (!high[rt]) {
                sum[rt] = 0;
            }
            return ;
        }
        int m = (l + r) >> 1;
        down(rt);
        fix(lson);
        fix(rson);
        up(rt);
    }
    void update(int L, int R, int d, int l, int r, int rt) {
        if (L <= l && r <= R) {
            low[rt] += d;
            high[rt] += d;
            late[rt] += d;
            if (low[rt]) {
                sum[rt] = r - l + 1;
            } else if (!high[rt]) {
                sum[rt] = 0;
            } else {
                fix(l, r, rt);
            }
            return ;
        }
        int m = (l + r) >> 1;
        down(rt);
        if (L <= m) update(L, R, d, lson);
        if (m < R) update(L, R, d, rson);
        up(rt);
    }
    int query() {
        fix(root);
        return sum[1];
    }
} segT;

struct Rect {
    int x1, y1;
    int x2, y2;
} ;
vector<Rect> rect;

void input(int n) {
    Rect buf;

    rect.clear();
    while (n--) {
        scanf("%d%d%d%d", &buf.x1, &buf.y1, &buf.x2, &buf.y2);
        if (buf.x1 > buf.x2) swap(buf.x1, buf.x2);
        if (buf.y1 > buf.y2) swap(buf.y1, buf.y2);
        rect.push_back(buf);
        swap(buf.x1, buf.x2);
        swap(buf.y1, buf.y2);
        rect.push_back(buf);
    }
}

bool cmpx(Rect a, Rect b) {
    return a.x1 < b.x1;
}

bool cmpy(Rect a, Rect b) {
    return a.y1 < b.y1;
}

int deal() {
    int last = 0, ret = 0;
    vector<Rect>::iterator ii;
    // cal x
    sort(rect.begin(), rect.end(), cmpx);
    segT.build(root);
    for (ii = rect.begin(); ii != rect.end(); ii++) {
        int y1 = (*ii).y1, y2 = (*ii).y2;
        if (y1 < y2) {
            segT.update(y1, y2 - 1, 1, root);
            ret += segT.query() - last;
        } else {
            segT.update(y2, y1 - 1, -1, root);
            ret += last - segT.query();
        }
        last = segT.query();
//        printf("y %d %d %d\n", last, y1, y2);
    }
    // cal y
    last = 0;
    sort(rect.begin(), rect.end(), cmpy);
    segT.build(root);
    for (ii = rect.begin(); ii != rect.end(); ii++) {
        int x1 = (*ii).x1, x2 = (*ii).x2;
        if (x1 < x2) {
            segT.update(x1, x2 - 1, 1, root);
            ret += segT.query() - last;
        } else {
            segT.update(x2, x1 - 1, -1, root);
            ret += last - segT.query();
        }
        last = segT.query();
//        printf("x %d %d %d\n", last, x1, x2);
    }

    return ret;
}

int main() {
    int n;

//    freopen("in", "r", stdin);
    while (~scanf("%d", &n)) {
        input(n);
        printf("%d\n", deal());
    }

    return 0;
}

UPD:

　　之前没看懂这道题的正确做法，然后被我yy了一个只是用线段树优化的代码，并没有起到线段树的真正作用。现在补上最新的，用线段树完美维护的代码，16ms通过。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;

#define _clr(x) memset(x, 0, sizeof(x))
#define REP(i, n) for (int i = 0; i < (n); i++)
#define REP_1(i, n) for (int i = 1; i <= (n); i++)

struct Side {
    int x1, x2, y, d;
    Side() {}
    Side(int _x1, int _x2, int _y, int _d) : x1(_x1), x2(_x2), y(_y), d(_d) {}
} x[11111], y[11111];

bool cmp(Side a, Side b) {
    if (a.y != b.y) return a.y < b.y;
    return a.d > b.d;
}

void input(int n) {
    int x1, y1, x2, y2;
    REP(i, n) {
        scanf("%d%d%d%d", &x1, &y1, &x2, &y2);
        if (x1 > x2) swap(x1, x2);
        if (y1 > y2) swap(y1, y2);
        x[i << 1] = Side(x1, x2 - 1, y1, 1);
        x[i << 1 | 1] = Side(x1, x2 - 1, y2, -1);
        y[i << 1] = Side(y1, y2 - 1, x1, 1);
        y[i << 1 | 1] = Side(y1, y2 - 1, x2, -1);
    }
    sort(x, x + (n << 1), cmp);
    sort(y, y + (n << 1), cmp);
}

const int N = 22222;
int sum[N << 2], cnt[N << 2];

#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define ROOT -10010, 10010, 1

void up(int rt, int len) {
    if (cnt[rt]) {
        sum[rt] = len;
    } else {
        sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
    }
}

void build(int l, int r, int rt) {
    sum[rt] = cnt[rt] = 0;
    if (l == r) return ;
    int m = (l + r) >> 1;
    build(lson);
    build(rson);
}

void update(int L, int R, int d, int l, int r, int rt) {
    if (L <= l && r <= R) {
        cnt[rt] += d;
        up(rt, r - l + 1);
        return ;
    }
    int m = (l + r) >> 1;
    if (L <= m) update(L, R, d, lson);
    if (m < R) update(L, R, d, rson);
    up(rt, r - l + 1);
}

int sweepLine(Side *s, int n) {
    n <<= 1;
    int ret = 0, cur = 0;
    build(ROOT);
    REP(i, n) {
        update(s[i].x1, s[i].x2, s[i].d, ROOT);
        ret += abs(cur - sum[1]);
        cur = sum[1];
    }
    return ret;
}

int main() {
//    freopen("in", "r", stdin);
    int n;
    while (~scanf("%d", &n)) {
        input(n);
        printf("%d\n", sweepLine(x, n) + sweepLine(y, n));
    }
    return 0;
}

　　这个代码里面，就算是成段操作，也不用将结果推倒儿子节点处。这样一来就可以实现返回历史历史状态的效果了。

——written by Lyon