标签:blog os io for ar 问题 cti div
有点像HDU 3642的强化版。给你N个矩形的坐标,问题平面上被k个不同的矩形覆盖的面积是多少。
当初HDU 3642 是直接一个一个手写的,这里的k虽然说只有10,合并过成一个一个手写也是相当蛋疼的,不过仔细想一下,不难推出一般性的关系,然后直接用循环搞就好了。不过我还是因为有个地方忘记初始化WA了2发,真是弱o(╯□╰)o
注意每个房子代表一个点,而我们要把他转化成线段,对坐标进行一些简单的变换即可。
#include <cstdio> #include <cstring> #include <iostream> #include <map> #include <set> #include <vector> #include <string> #include <queue> #include <deque> #include <bitset> #include <list> #include <cstdlib> #include <climits> #include <cmath> #include <ctime> #include <algorithm> #include <stack> #include <sstream> #include <numeric> #include <fstream> #include <functional> using namespace std; #define MP make_pair #define PB push_back #define lson rt << 1,l,mid #define rson rt << 1 | 1,mid + 1,r typedef long long LL; typedef unsigned long long ULL; typedef vector<int> VI; typedef pair<int,int> pii; const int INF = INT_MAX / 3; const double eps = 1e-8; const LL LINF = 1e17; const double DINF = 1e60; const int maxn = 3e4 + 20; struct Seg { int x,l,r,cover; Seg(int x,int l,int r,int cover): x(x),l(l),r(r),cover(cover) {} bool operator < (const Seg &s) const { return x < s.x; } }; int cnt[maxn << 4],sum[maxn << 4][11]; VI numy; vector<Seg> seg; int n,k; void pushup(int rt,int l,int r) { int lc = rt << 1, rc = rt << 1 | 1; if(cnt[rt] >= k) { sum[rt][k] = numy[r + 1] - numy[l]; for(int i = 1;i < k;i++) sum[rt][i] = 0; } else if(cnt[rt] == 0) { for(int i = 1;i <= k;i++) sum[rt][i] = sum[lc][i] + sum[rc][i]; } else { int nowsum = 0; for(int i = k;i >= 1;i--) { if(i < cnt[rt]) sum[rt][i] = 0; else if(i == cnt[rt]) sum[rt][i] = numy[r + 1] - numy[l] - nowsum; else { sum[rt][i] = 0; for(int j = 1;j <= i;j++) if(j + cnt[rt] >= i) { if(i == k) sum[rt][i] += sum[lc][j] + sum[rc][j]; else if(j + cnt[rt] == i) sum[rt][i] += sum[lc][j] + sum[rc][j]; } nowsum += sum[rt][i]; } } } } void update(int rt,int l,int r,int ql,int qr,int Val) { if(ql <= l && qr >= r) { cnt[rt] += Val; pushup(rt,l,r); } else { int mid = (l + r) >> 1; if(ql <= mid) update(lson,ql,qr,Val); if(qr > mid) update(rson,ql,qr,Val); pushup(rt,l,r); } } int getID(int Val) { return lower_bound(numy.begin(),numy.end(),Val) - numy.begin(); } int main() { int T,x1,y1,x2,y2; scanf("%d",&T); for(int kase = 1;kase <= T;kase++) { scanf("%d%d",&n,&k); numy.clear(); seg.clear(); for(int i = 0;i < n;i++) { scanf("%d%d%d%d",&x1,&y1,&x2,&y2); seg.PB(Seg(x1,y1,y2 + 1,1)); seg.PB(Seg(x2 + 1,y1,y2 + 1,-1)); numy.PB(y1); numy.PB(y2 + 1); } sort(numy.begin(),numy.end()); sort(seg.begin(),seg.end()); numy.erase(unique(numy.begin(),numy.end()),numy.end()); int ks = seg.size(), ky = numy.size(); LL ans = 0; for(int i = 0;i < ks;i++) { int ql = getID(seg[i].l), qr = getID(seg[i].r) - 1; update(1,0,ky - 1,ql,qr,seg[i].cover); if(i < ks - 1) { ans += 1LL * (seg[i + 1].x - seg[i].x) * sum[1][k]; } } cout << "Case " << kase << ": " << ans << endl; } return 0; }
UVA 11983 Weird Advertisement 线段树+离散化+扫描线,布布扣,bubuko.com
UVA 11983 Weird Advertisement 线段树+离散化+扫描线
标签:blog os io for ar 问题 cti div
原文地址:http://www.cnblogs.com/rolight/p/3924584.html