Cubes:
我们先将每个正方形的光路投影找出来（在y轴上的截距所构成的区间）
然后，我们再考虑按照光线射到的先后顺序加入每个方格。
再用线段树统计这个投影区间内的最小高度的光线。这样我们就能够算出每个点被照亮的高度。
复杂度O(n2logn+).
注意：
在计算区间的时候选择前闭后开区间，这是因为我们要满足两个条件，
1.只通过一个点的光线不算照亮
2.假设先后修改（2,3），（1,2），（4,5）那么现在的（2,3）区间不能是(1,2)或(4,5)中间的一个.
代码很丑,凑合着看吧.(数据似乎没有和坐标轴平行的光,我就没管了.....)
1 #include<cstdlib>
2 #include<cstdio>
3 #include<algorithm>
4 #include<cstring>
5 #include<iostream>
6 #include<map>
7 using namespace std;
8 const int maxn = 510, maxs = maxn * maxn * 5,inf = 0x3f3f3f3f;
9 const int dx[] = {0,0,0,1,1},
10 dy[] = {0,0,1,0,1};
11 int ct;
12 struct SEG{
13 int sg[maxs * 3],tag[maxs * 3];
14 void down(int x){
15 if(tag[x]){
16 sg[x << 1] = max(sg[x << 1], tag[x]);
17 sg[x << 1|1] = max(sg[x << 1|1], tag[x]);
18 tag[x << 1] = max(tag[x << 1], tag[x]);
19 tag[x <<1|1] = max(tag[x << 1|1], tag[x]);
20 tag[x] = 0;
21 }
22 }
23 int qer(int x,int l,int r,int nl,int nr){
24 down(x);
25 if(l == nl && nr == r) return sg[x];
26 int mid = (nl + nr) >> 1;
27 if(r <= mid) return qer(x << 1, l, r, nl, mid);
28 else if(l > mid) return qer(x << 1|1, l, r, mid + 1, nr);
29 else{
30 int t1 = qer(x << 1, l, mid, nl, mid);
31 int t2 = qer(x << 1|1, mid + 1, r, mid+1, nr);
32 return min(t1,t2);
33 }
34 }
35 void mod(int x,int l,int r,int nl,int nr,int val){
36 down(x);
37 if(l == nl && nr == r){
38 sg[x] = max(sg[x],val), tag[x] = max(tag[x],val);
39 return;
40 }
41 int mid = (nl + nr) >> 1;
42 if(r <= mid){
43 mod(x << 1, l, r, nl, mid, val);
44 }
45 else if(l > mid){
46 mod(x << 1|1, l, r, mid + 1, nr, val);
47 }
48 else{
49 mod(x << 1, l, mid, nl, mid, val);
50 mod(x << 1|1, mid + 1, r, mid + 1, nr, val);
51 }
52 sg[x] = min(sg[x << 1], sg[x << 1|1]);
53 }
54
55 }seg;
56 int n,vx,vy;
57 int tot;
58 struct grid{
59 int x,y;
60 long long height;
61 }g[maxn * maxn + 100];
62 int cmp(grid x, grid y){
63 if(x.x != y.x) return x.x > y.x;
64 return x.y > y.y;
65 }
66 void trans(){
67 if(vx > 0){
68 vx = -vx;
69 for(int i = 1; i <= tot; ++i) g[i].x = n - g[i].x - 1;
70 }
71 if(vy > 0){
72 vy = -vy;
73 for(int i = 1; i <= tot; ++i) g[i].y = n - g[i].y - 1;
74 }
75 if(vx == 0 && vy != 0){
76 swap(vx,vy);
77 for(int i = 1; i <= tot; ++i){
78 swap(g[i].x, g[i].y);
79 g[i].y = n - g[i].y - 1;
80 }
81 }
82 }
83 struct I{
84 long long it;
85 int rank;
86 }data[maxn][maxn];
87 map<long long, int>v;
88 map<long long, int>::iterator itr;
89 void inter(int x,int y){
90 long long t = ((long double) y - vy * x / (long double)vx) * (long long)1e9;
91 data[x][y] = (I){t,0};
92 v[t] = 0;
93 }
94 void calc(){
95 for(int i = 0; i <= n; ++i)
96 for(int j = 0; j <= n; ++j)
97 inter(i,j);
98 for(itr = v.begin(); itr != v.end(); ++itr)
99 itr->second = ++ct;
100 for(int i = 0; i <= n; ++i)
101 for(int j = 0; j <= n; ++j){
102 data[i][j].rank = v[data[i][j].it];
103 }
104 }
105 long long ans;
106 void ins(){
107 sort(g + 1, g + tot + 1, cmp);
108 for(int i = 1; i <= tot; ++i){
109 int up = -inf, down = inf;
110 for(int j = 1; j <= 4; ++j){
111 int tx = g[i].x + dx[j], ty = g[i].y + dy[j];
112 up = max(up, data[tx][ty].rank);
113 down = min(down, data[tx][ty].rank);
114 }
115 int r = up - 1, l = down;
116 if(l <= r && up >= 0){
117 int t = seg.qer(1, l , r, 1, ct);
118 ans += max(g[i].height - t, 0LL);
119 seg.mod(1, l, r, 1, ct, g[i].height);
120 }
121 }
122 }
123 void spj(){
124 long long ans = 0;
125 sort(g + 1, g + tot + 1, cmp);
126 for(int i = 1; i <= tot; ++i){
127 int x = g[i].x, y = g[i].y;
128 data[x][y].it = max(data[x][y].it, data[x+1][y].it);
129 ans += max(g[i].height - data[x][y].it, 0LL);
130 data[x][y].it = max(data[x][y].it, g[i].height);
131 }
132 cout << ans;
133 }
134 int main()
135 {
136 freopen("cubes.in","r",stdin);
137 freopen("cubes.out","w",stdout);
138 ios::sync_with_stdio(false);
139 cin >> n >> vx >> vy;
140 for(int i = 1; i <= n; ++i)
141 for(int j = 1; j <= n; ++j){
142 int t; cin >> t;
143 g[++tot] = (grid){i-1,j-1,t};
144 }
145 trans();
146 calc();
147 ins();
148 cout << ans;
149 return 0;
150 }
时间： 2024-10-12 17:13:16