BZOJ3888 [Usaco2015 Jan]Stampede

我们只要把每头牛开始遮挡视线和结束遮挡视线的时间点都搞出来就好= =

再按照y轴排序。。。然后变成线段覆盖了。。线段树搞一下就好了？

 /**************************************************************
     Problem: 3888
     User: rausen
     Language: C++
     Result: Accepted
     Time:1204 ms
     Memory:22684 kb
 ****************************************************************/

 #include <cstdio>
 #include <algorithm>

 using namespace std;
 const int N = 1e5 + ;

 struct data {
     int x, y, c;
     int st, ed;

     inline void read() {
         scanf("%d%d%d", &x, &y, &c);
         x *= -, st = c * (x - ), ed = st + c;
     }

     inline bool operator < (const data &p) const {
         return y > p.y;
     }
 } a[N];

 struct seg_node {
     seg_node *ls, *rs;
     int v;
 } *seg_null, *seg_root, mempool[N << ], *cnt_seg = mempool;

 int n, ans;
 int v[N];
 int t[N << ], tot;

 #define Ls p -> ls
 #define Rs p -> rs
 #define V p -> v
 #define mid (l + r >> 1)
 inline void seg_get_null(seg_node *&p) {
     p = cnt_seg, Ls = Rs = p;
 }

 inline void seg_new(seg_node *&p) {
     p = ++cnt_seg, *p = *seg_null;
 }

 inline void seg_push(seg_node *p) {
     if (V) {
         Ls -> v = Rs -> v = V;
         V = ;
     }
 }

 inline void seg_build(seg_node *&p, int l, int r) {
     seg_new(p);
     if (l == r) return;
     seg_build(Ls, l, mid), seg_build(Rs, mid + , r);
 }

 inline void seg_modify(seg_node *p, int l, int r, int L, int R, int v) {
     if (L <= l && r <= R) {
         V = v;
         return;
     }
     seg_push(p);
     if (L <= mid) seg_modify(Ls, l, mid, L, R, v);
     if (mid < R) seg_modify(Rs, mid + , r, L, R, v);
 }

 inline int seg_query(seg_node *p, int l, int r, int pos) {
     if (l == r) return V;
     seg_push(p);
     if (pos <= mid) return seg_query(Ls, l, mid, pos);
     else return seg_query(Rs, mid + , r, pos);
 }
 #undef Ls
 #undef Rs
 #undef V
 #undef mid

 int main() {
     int i;
     scanf("%d", &n);
     for (i = ; i <= n; ++i)
         a[i].read();
     sort(a + , a + n + );
     for (i = ; i <= n; ++i)
         t[i *  - ] = a[i].st, t[i * ] = a[i].ed;
     sort(t + , t +  * n + );
     tot = unique(t + , t +  * n + ) - t;
     for (i = ; i <= n; ++i) {
         a[i].st = lower_bound(t + , t + tot + , a[i].st) - t;
         a[i].ed = lower_bound(t + , t + tot + , a[i].ed) - t;
         a[i].st *= , a[i].ed *= ;
     }
     tot *= ;
     seg_get_null(seg_null);
     seg_build(seg_root = seg_null, , tot);
     for (i = ; i <= n; ++i)
         seg_modify(seg_root, , tot, a[i].st, a[i].ed, i);
     for (i = ; i <= tot; ++i)
         v[seg_query(seg_root, , tot, i)] = ;
     for (ans = , i = ; i <= n; ++i)
         ans += v[i];
     printf("%d\n", ans);
     return ;
 }