UVA11983 - Weird Advertisement(扫描线)

UVA11983 - Weird Advertisement(扫描线)

题目链接

题目大意：给你n个覆盖矩形，问哪些整数点是被覆盖了k次。

题目大意：这题和hdu1542是一个题型。可是这题求的是覆盖k次的点。所以pushup里面就要改一下。详细的看代码把。大概的意思就是每次都是通过以下的两个孩子节点的覆盖信息更新父节点的覆盖信息。然后这题也是须要离散化建树。比較须要注意的是这题和hdu1542不一样的地方是边界，由于那题是求面积。边界并不须要考虑。而这题是求点的个数，边界上的点也是要算的。建树的时候考虑叶子节点范围的时候就要包含边界上的点。

代码：

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

using namespace std;

#define lson(x) (x<<1)
#define rson(x) ((x<<1) | 1)

const int maxn = 60005;
typedef long long ll;

struct Line {

    int x, y1, y2, flag;
    Line (int x, int y1, int y2, int flag) {

        this->x = x;
        this->y1 = y1;
        this->y2 = y2;
        this->flag = flag;
    }

    bool operator < (const Line & a) const {
        return x < a.x;
    }
};

struct Node {

    int l, r, add;
    ll s[12];
    void set (int l, int r, int add) {

        this->l = l;
        this->r = r;
        this->add = add;
    }
}node[maxn * 4];

vector<int> pos;
vector<Line> L;
int n, k;

void pushup(int u) {

    for (int i = 0; i <= k; i++)
        node[u].s[i] = 0L;

    int t;
    if (node[u].l == node[u].r) {
        t = min(k, node[u].add);
        node[u].s[t] = pos[node[u].l + 1] - pos[node[u].l];
    } else {
        for (int i = 0; i <= k; i++) {
            t = min (i + node[u].add, k);
            node[u].s[t] += node[lson(u)].s[i] + node[rson(u)].s[i];
        }
    }
}

void build (int u, int l, int r) {

    node[u].set(l, r, 0);

    if (l == r) {
        pushup(u);
        return;
    }

    int m = (l + r)>>1;
    build(lson(u), l, m);
    build(rson(u), m + 1, r);
    pushup(u);
}

void update (int u, int l, int r , int v) {

    if (l <= node[u].l && r >= node[u].r) {
        node[u].add += v;
        pushup(u);
        return;
    }

    int m = (node[u].l + node[u].r)>>1;
    if (l <= m)
        update (lson(u), l, r, v);
    if (r > m)
        update (rson(u), l, r, v);
    pushup(u);
}

void init () {

    int x1, y1, x2, y2;
    scanf ("%d%d", &n, &k);
    L.clear();
    pos.clear();

    for (int i = 0; i < n; i++) {

        scanf ("%d%d%d%d", &x1, &y1, &x2, &y2);

        L.push_back(Line(x1, y1, y2 + 1, 1));
        L.push_back(Line(x2 + 1, y1, y2 + 1, -1));
        pos.push_back(y1);
        pos.push_back(y2 + 1);
    }

    sort(L.begin(), L.end());
    sort(pos.begin(), pos.end());
    pos.erase (unique(pos.begin(), pos.end()), pos.end());

    build(1, 0, (int)pos.size() - 1);
}

ll solve () {

    init();
    ll ans = 0;

    int l, r;
    for (int i = 0; i < L.size() - 1; i++) {

        l = lower_bound(pos.begin(), pos.end(), L[i].y1) - pos.begin();
        r = lower_bound(pos.begin(), pos.end(), L[i].y2) - pos.begin();

        update (1, l, r - 1, L[i].flag);
        ans += node[1].s[k] * (L[i + 1].x - L[i].x);
    }
    return ans;
}

int main () {

    int T;
    scanf ("%d", &T);

    for (int cas = 1; cas <= T; cas++)
        printf ("Case %d: %lld\n", cas, solve());
    return 0;
}

版权声明：本文博客原创文章，博客，未经同意，不得转载。