拍照

树状数组。求出静止状态下，每个点能看到多少个向右开的船c1[i]，多少个向左开的船c2[i]。

max{c1[i],   max{ c2[j], (i <= j) } }即为答案。

注意要离散化，否则会Tle。

 1 #include <bits/stdc++.h>
 2 typedef long long ll;
 3 using namespace std;
 4
 5 const int maxn =2e4;
 6 //c1 向右开的船
 7 int c1[maxn<<2], c2[maxn<<2];
 8 int s1[maxn<<2], s2[maxn<<2];
 9
10 int lowbit(int x){ return x&-x;}
11 void add(int x, int d, int* c){//c[x]++;
12     while(x <= (maxn<<2)){
13         c[x] += d;
14         x += lowbit(x);
15     }
16 }
17 void Add(int l, int r, int* c){
18     add(l, 1, c);
19     add(r+1, -1, c);
20 }
21 int sum(int x, int* c){
22     int ret = 0;
23     while(x > 0){
24         ret += c[x];
25         x -= (x&-x);
26     }
27     return ret;
28 }
29
30 int n;
31 int ope[maxn], tot;
32 struct p{
33     int l, r, d;
34     p(){}
35     p(int l, int r, int d):l(l), r(r), d(d){}
36 };
37 p pp[maxn];
38
39 int main(){
40     int t, ca = 1;scanf("%d", &t);
41     while(t--){
42         scanf("%d", &n);
43         memset(c1, 0, sizeof(c1));
44         memset(c2, 0, sizeof(c2));
45         int l, r, x, y, z, d;
46
47         tot = 0;
48         for(int i = 0; i < n; i++){
49             scanf("%d%d%d%d", &x, &y, &z, &d);
50             l = y-z+maxn, r = x+z+maxn;
51             ope[tot++] = l, ope[tot++] = r;
52             pp[i] = p(l, r, d);
53         }
54
55         sort(ope, ope+tot);
56         //tot = unique(ope, ope+tot)-ope;
57         for(int i = 0; i < n; i++){
58             int l = lower_bound(ope, ope+tot, pp[i].l)-ope+1;
59             int r = lower_bound(ope, ope+tot, pp[i].r)-ope+1;
60             int d = pp[i].d;
61             if(l <= r)
62                 Add(l, r, (d == 1? c1 : c2));
63         }
64
65         for(int i = 0; i < (maxn<<2); i++)
66             s1[i] = sum(i, c1);
67         for(int i = 0; i < (maxn<<2); i++)
68             s2[i] = sum(i, c2);
69
70         int ans = 0;
71         for(int i = (maxn<<2)-2; i > 0; i--){
72             s2[i] = max(s2[i], s2[i+1]);
73             ans = max(ans, s1[i]+s2[i]);
74         }
75         printf("Case #%d:\n%d\n", ca++, ans);
76     }
77     return 0;
78 }