uva 11983(扫描线)

来源：互联网发布：ct young无网络连接编辑：程序博客网时间：2024/05/21 08:53

题意：有n个矩形，给出每个矩形的左下角坐标和右上角坐标，问所有矩形重叠k次以上的面积并。
题解：因为k最大是10，直接用线段树维护重叠1到k-1次的面积并和大于等于k的面积并，pushup函数更新每个重叠次数。

#include <cstdio>#include <cstring>#include <algorithm>#include <map>#include <vector>#define ll long longusing namespace std;const int N = 60005;struct Line {    int lx, rx, h, flag;    Line(int a, int b, int c, int d): lx(a), rx(b), h(c), flag(d) {}    bool operator < (const Line& a) const { return h < a.h; }};int n, m, flag[N << 2];ll tree[12][N << 2];vector<int> a;vector<Line> line;map<int, int> mp;void pushup(int k, int left, int right) {    if (flag[k] >= m) {        tree[m][k] = tree[0][k];        for (int i = 1; i < m; i++)            tree[i][k] = 0;    }    else if (flag[k] > 0) {        if (left + 1 == right) {            for (int i = 1; i <= m; i++)                tree[i][k] = 0;            tree[flag[k]][k] = tree[0][k];        }        else {            tree[m][k] = 0;            for (int i = m - flag[k]; i <= m; i++)                tree[m][k] += tree[i][k * 2] + tree[i][k * 2 + 1];            for (int i = 1; i + flag[k] < m; i++)                tree[i + flag[k]][k] = tree[i][k * 2] + tree[i][k * 2 + 1];            tree[flag[k]][k] = tree[0][k];            for (int i = flag[k] + 1; i <= m; i++)                tree[flag[k]][k] -= tree[i][k];            for (int i = 1; i < flag[k]; i++)                tree[i][k] = 0;        }    }    else {        if (left + 1 == right) {            for (int i = 1; i <= m; i++)                tree[i][k] = 0;        }        else {            for (int i = 1; i <= m; i++)                tree[i][k] = tree[i][k * 2] + tree[i][k * 2 + 1];        }    }}void build(int k, int left, int right) {    flag[k] = 0;    tree[0][k] = a[right] - a[left];    for (int i = 1; i <= m; i++)        tree[i][k] = 0;    if (left + 1 != right) {        int mid = (left + right) / 2;        build(k * 2, left, mid);        build(k * 2 + 1, mid, right);    }}void modify(int k, int left, int right, int l, int r, int v) {    if (l <= left && right <= r) {        flag[k] += v;        pushup(k, left, right);        return;    }    int mid = (left + right) / 2;    if (l < mid)        modify(k * 2, left, mid, l, r, v);    if (r > mid)        modify(k * 2 + 1, mid, right, l, r, v);    pushup(k, left, right);}int main() {    int t, cas = 1;    scanf("%d", &t);    while (t--) {        a.clear(), line.clear(), mp.clear();        scanf("%d%d", &n, &m);        int x1, y1, x2, y2;        for (int i = 1; i <= n; i++) {            scanf("%d%d%d%d", &x1, &y1, &x2, &y2);            x2++, y2++;            line.push_back(Line(x1, x2, y1, 1));            line.push_back(Line(x1, x2, y2, -1));            a.push_back(x1);            a.push_back(x2);        }        sort(line.begin(), line.end());        sort(a.begin(), a.end());        a.erase(unique(a.begin(), a.end()), a.end());        int sz = a.size(), sz2 = line.size();        for (int i = 0; i < sz; i++)            mp[a[i]] = i;        build(1, 0, sz - 1);        ll res = 0;        for (int i = 0; i < sz2; i++) {            if (i != 0)                res += tree[m][1] * (ll)(line[i].h - line[i - 1].h);            modify(1, 0, sz - 1, mp[line[i].lx], mp[line[i].rx], line[i].flag);        }        printf("Case %d: %lld\n", cas++, res);    }    return 0;}

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