题目大意
给出一些奶牛，一个人在原点观察，牛和牛之间又互相遮挡的关系，给出每头牛的运行方式和位置，问这个人最终会看到多少头牛。
思路
知道了运行方式，我们就知道这头牛在什么时间段会遮挡住人的视线，然后从高到低弄个东西维护一下覆盖什么的，这个题就变成了POJ的Mayor’s posters。
注意下时间点和时间段的区别就行了。
CODE
#define _CRT_SECURE_NO_WARNINGS
#include
#include
#include
#include
#define MAX 100010 using namespace std; #define LEFT (pos << 1) #define RIGHT (pos << 1|1) struct Cow{ int x,y,c; int st,ed; bool operator <(const Cow &a)const { return y > a.y; } void Read() { scanf("%d%d%d", &x, &y, &c); x *= -1; st = c * (x - 1); ed = st + c; } }src[MAX]; int cows; pair
xx[MAX << 1]; int cnt,t; int tree[MAX << 4]; inline void PushDown(int pos) { if(tree[pos]) { tree[LEFT] = tree[pos]; tree[RIGHT] = tree[pos]; tree[pos] = 0; } } void Modify(int l, int r, int x, int y, int c, int pos) { if(l == x && y == r) { tree[pos] = c; return ; } PushDown(pos); int mid = (l + r) >> 1; if(y <= mid) Modify(l, mid, x, y, c, LEFT); else if(x > mid) Modify(mid + 1, r, x, y, c, RIGHT); else { Modify(l, mid, x, mid, c, LEFT); Modify(mid + 1, r, mid + 1, y, c, RIGHT); } } inline int Ask(int l, int r, int x, int pos) { if(l == r) return tree[pos]; PushDown(pos); int mid = (l + r) >> 1; if(x <= mid) return Ask(l, mid, x, LEFT); return Ask(mid + 1, r, x, RIGHT); } bool v[MAX]; int main() { cin >> cows; for(int i = 1; i <= cows; ++i) src[i].Read(); sort(src + 1, src + cows + 1); for(int i = 1; i <= cows; ++i) { xx[++cnt] = make_pair(src[i].st, &src[i].st); xx[++cnt] = make_pair(src[i].ed, &src[i].ed); } sort(xx + 1, xx + cnt + 1); for(int i = 1; i <= cnt; ++i) { if(i == 1 || xx[i].first != xx[i - 1].first) t += 2; *xx[i].second = t; } for(int i = 1; i <= cows; ++i) Modify(1, t, src[i].st, src[i].ed , i, 1); for(int i = 1; i <= t; ++i) v[Ask(1, t, i, 1)] = true; int ans = 0; for(int i = 1; i <= cows; ++i) ans += v[i]; cout << ans << endl; return 0; }