BZOJ 3888 Usaco 2015 Jan Stampede 模拟

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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 <cstdio>#include <cstring>#include <iostream>#include <algorithm>#define MAX 100010using 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<int,int *> 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;}

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