uva 1232 - SKYLINE(线段树)

题目链接：uva 1232 - SKYLINE

题目大意：就是n次修改维护最大值，如果一个位置的最大值被修改了，覆盖值就+1，问说总的覆盖值。

解题思路：线段树的区间修改，每次修改后返回修改到的区间长度即可。

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

using namespace std;
const int maxn = 100000;
#define lson(x) ((x)<<1)
#define rson(x) (((x)<<1)+1)

struct Node {
    int l, r, sign, high;
    void set (int l, int r, int sign, int high) {
        this->l = l;
        this->r = r;
        this->sign = sign;
        this->high = high;
    }
}node[maxn*4+5];

void pushup(int u) {
    node[u].l = node[lson(u)].l;
    node[u].r = node[rson(u)].r;

    if (node[lson(u)].sign && node[rson(u)].sign
            && node[lson(u)].high == node[rson(u)].high) {
        node[u].sign = 1;
        node[u].high = node[lson(u)].high;
    } else {
        node[u].sign = 0;
        node[u].high = 0;
    }
}

void build_segTree (int u, int l, int r) {
    if (l == r) {
        node[u].set(l, r, 1, 0);
        return;
    }

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

void pushdown (int u) {
    node[lson(u)].sign = node[rson(u)].sign = 1;
    node[lson(u)].high = node[rson(u)].high = node[u].high;
    node[u].sign = node[u].high = 0;
}

int insert (int u, int l, int r, int h) {
    if (node[u].sign && node[u].high > h)
        return 0;

    if (l <= node[u].l && node[u].r <= r && node[u].sign) {
        node[u].high = h;
        return node[u].r - node[u].l + 1;
    }

    if (node[u].sign)
        pushdown(u);

    int mid = (node[u].l + node[u].r) / 2;
    int ret = 0;

    if (mid >= l)
        ret += insert(lson(u), l, r, h);
    if (mid < r)
        ret += insert(rson(u), l, r, h);
    pushup(u);
    return ret;
}

int main () {
    int cas, n;
    scanf("%d", &cas);
    while (cas--) {
        scanf("%d", &n);
        build_segTree(1, 1, maxn);

        int ans = 0, x, y, h;
        for (int i = 0; i < n; i++) {
            scanf("%d%d%d", &x, &y, &h);
            ans += insert(1, x+1, y, h);
        }
        printf("%d\n", ans);
    }
    return 0;
}